hello everyone so this is the first installment of my three or four parts lecture series for functions limits and uh continuity and these three words function limits continuity are very important on the study of calculus and we shall start our lecture by defining a functions and it's a very common knowledge for us that the function assigns rule to describe how a certain quantity depends on the values of another variables and in particular the equation of the line y equals m x plus b sets the relationship between the variables x and y and this relationship can be thought as a correspondence from a set x a real number x and to accept y a real number of y so technically we define the function to be the set of ordered pairs x y in which no two distinct ordered pairs have the same first number and this real number x is what we call the domain and this set of real number y is called the range and um small letters f g and h denotes a function unless stated otherwise and when written or when expressed in terms of the variable x we let f of x g of x h of x to denote a function so for instance if we let f be defined by f of x equals to x minus 1 then it can be observed that f is a function since when x is replaced by any number or any real number then we can have exactly one value of y okay for instance when x is negative three y equals negative seven when x is negative two y is negative five when x is negative three or negative one uh f of x is negative three when x is zero f of x is negative one when x is one we have one when x is two we have three and finally when x is three we have five and this set containing negative three negative two negative one zero one two and three is the domain and negative 7 negative 5 negative 3 negative 1 1 3 and 5 is called the range and you will notice that not two or more ordered pairs have the same first number so the the relation f defined by f of x equals 2x minus 1 can be thought of can be thought as a functions and if we let f be defined by the circle x squared plus y squared equals 9 then you will notice that when x is zero when x is replaced by zero you will have y equals three and negative three so we have the set of ordered pairs zero three and zero negative three and based on our definitions uh a function is a set of ordered pairs we're in no two or more ordered pairs of the same first number so we have zero and another zero so this is zero three this is zero negative three so x squared plus y squared equals 9 cannot be thought as a function but it's a relation except rules for the variables x and y similarly the ordered pairs 3 0 negative 3 0 1 2 square root of 2 and 1 negative the square root of 2 can also be obtained from the given circle okay and though how do we define or how do we identify whether a relation is a function or not okay when a graph is given then we can easily identify a relation whether it's a function or not through a vertical line test so here is the idea so what are we going to do is to draw vertical lines that will pass through the graph of the uh relation okay and if the vertical line passes through exactly two or more points on the curve then that is not a function and you will notice that when i draw vertical lines over the circle x squared plus y squared equals nine uh the vertical line passes through exactly two points so that confirms that the circle is not a function however uh if we uh draw vertical lines over this curve then you will notice that there is exactly one point on the curve by which the the vertical line passes through so which mean this uh relation or this curve is a function and another way to do it is by mapping right so let us consider the the the relation f of x equals to x minus 1 and we know that when x is negative 3 y f of x is negative 7 and negative 2 maps to negative 5 negative 1 maps to negative 3. zero maps to negative ones one maps to one two maps to three and three maps to five and you will notice that for every uh value of x there is exactly one value of y mapped to it so therefore we have a one to one correspondence or one-to-one mapping so therefore we can talk a one-to-one correspondence as a function okay going back with our circle x squared plus y squared equals nine uh earlier we have identified that the circle is not a function but a relation of course so by mapping so you will observe that negative three maps to zero when x is zero it will be mapped to negative three and positive three when x is 1 then we have negative 2 square root of 2 into square root of 2 and when x is 3 we have 0 so you will notice that there is exactly one value of x mapped to 2 values of our y okay uh there is uh one is mapped to two values of y and therefore we have one too many uh correspondence and therefore the one to minimapping cannot be taught as a function and another one we have here g of x equals x squared plus 1 and then you will notice that when x is negative 3 y is negative 10 or g of x is negative 10 or positive 10 sorry and when x is negative two we have five negative one mops to two zero maps to one okay so the idea is i am just replacing x by this numbers and then we obtain g of x so when x is when x is replaced by 1 so 1 squared plus 1 so we have 2 when x is 2 we have 5 and when x is 3 g of x is 10 and you will notice that there are two values of x mapped to only one value of j of x again uh say three and negative three are both uh mapped to ten and therefore we have many to one uh mapping or many to one correspondence and the main e-to-one mapping or mainly to one correspondence can be taught as a function right how to evaluate a function one easiest way to evaluate the function is by a straightforward substitution for instance if we let f be a function defined by f of x equals x squared minus two x plus three and if we want to get this value say f of two f of one half f of negative one f of two x and f of x plus h and then we can do a straightforward substitution for f of 2 you will just be replacing x by 2 so you have 2 square minus 2 times 2 plus 3 and you obtain 3 so f of 2 equals 3. for letter b if we replace x by one half so we have f of one half equals one half squared minus two times one half plus three so one half squared is one fourth and negative two times one half is negative one and then plus three and then you have 9 4 uh to be our f of one half and then f of two x so sorry f of negative one so again you replace x by negative one so negative one square minus two times negative one plus three and that will give you positive six this is one and this is two so one plus two is three plus three you have six for f of two x so here you replace x by two x so you have the quantity two x squared minus two times two x plus three then two x squared is four x squared minus two times two x is four x plus three so f of two x is given by four x squared minus four x plus three and then finally if we want to evaluate f of x plus h so this time we will be replacing x by x plus h so you have the quantity x plus h squared minus two times the quantity x plus h plus three and then simplifying this algebraically we have x squared plus two x h plus h squared that's the uh the square of x plus h the square of the first plus twice the product of the first and the second square of the second and then you introduce negative two each term here so you have negative two x minus two 2h plus 3 and i think we can no longer combine anything here so the final answer would have been equal to x squared plus 2xh plus h squared minus 2x minus 2h plus 3. okay how to find the domain in the range again the dominant the range of the function is also an important characteristic of a function again the domain is the set of the real number x and the range is the set of the real number y okay and then by definition the set of all admissible values of x is called the domain and this and then the set of all the permissible uh values of y is called the range of the function and say we have a function f of x equals two x plus three so what are you going to do is to to identify what are those values of x that can be used to replace x here what are those real numbers that can be used to replace x here can x be replaced by positive number can x be replaced by zero can x be replaced by negative numbers such that the result is still defined okay and it can be thought that [Laughter] you can replace x by any number on the uh set of real numbers so the domain of this function is the set containing x such that x is an element of real numbers or in interval notation we use this negative the open interval negative infinity to positive infinity now for the range what are you going to ask for is after replacing x by any number on this interval what happened to your f of x if i replace x by negative number what happened there f of x if i replace x by zero how do you characterize f of x if x is replaced by positive number then how do you describe your f of x and uh replacing x by a number on this interval then we can say that the range of this function is the set of real numbers or all reals or in interval notation that is negative infinity to positive infinity okay okay say if we let g be a function defined by g of x equals square root of x and then we have here a square root of x and then we know that if a number is written inside the square root then this number x cannot be negative okay so that's an important consideration so we know that the domain of this function is the set of all non-negative real numbers or the set the the the the set containing zero to positive infinity and in in interval notation this is read as the closed open interval of 0 to positive infinity and when x is replaced by any number on this interval we say that the range of this function is the set containing 0 to positive infinity including zero okay if we want to teach this curve then we can have this so your g of x equals square root of x is this represented by this uh graph or by this curve okay and you will notice that the dominant this function is from zero to positive infinity and then the range is from zero to positive infinity okay another one say let that be a function defined by f of x equals 2 over x and if we want to find its dominant range again this is a rational function of the form p of x over q of x so we know that uh whenever you were given a fraction the denominator should not be equal to zero why because when the denominator is zero then the the function f of x is undefined because we are thought that dividing by zero is or is undefined okay so which means we will be excluding zero uh on the set of our domain so the domain of this function is the the interval negative infinity to zero union the interval zero to positive infinity or this is the same as the set of real numbers except zero okay now for the range so we have uh negative infinity to zero union zero to positive infinity all right so this is how we draw the curve so you know that x cannot be zero okay and y cannot be zero okay so that is the reason why the the curve uh y uh f of x equals two over x is asymptotic with respect to the x-axis and also asymptotic with respect to the y-axis when we say asymptotic it means that the curve or the graph of that function will just go nearer and near to x-axis but will never touch the x-axis so that is what i mean when i say asymptotic with respect to the x-axis so when it is asymptotic with respect to the y-axis the behavior of the curve uh gets nearer and nearer closer and closer to the y-axis but will never cross the y-axis okay and that is what you called the horizontal asymptote okay and the y-axis is the vertical asymptote for this particular function f of x equals two over x okay another one say we modify the the given uh function we now let f be defined by f of x equals 1 over x squared so this time the denominator x squared so how we may replace x so can we replace x by negative numbers obviously yes because when x is replaced by a negative number the result is still defined now can we replace x by positive number of course yes because when x is replaced by positive number still the function f 1 over x squared is still defined now when x is replaced by 0 so this will become 1 over 0 so the function is no longer defined at x equals 0. so we want to trace it so as you notice that when x is uh approaching positive a very very large number the value of y diminished or the value of y decreases and it gets closer and closer to zero and as x approaches zero as x as the value of x gets closer and closer to zero okay the value of your f of x gets uh larger and larger approaching positive infinity and since y cannot be negative because if we replace x by any number here positive x or negative number then still the result will become positive because of this x squared because of that square so the other part of the curve is this okay and as x approaches a very large number on the left side again y approaches x and as x gets closer and closer to zero from the left your y gets larger and larger approaching positive infinity and again this is your horizontal asymptote and then this one is your vertical asymptote okay and then the dimension uh the domain of this function as you can notice so that's the set of real numbers except zero or the interval negative infinity to zero union the open interval zero to positive infinity and as you can notice there's no trace of the curve on this portion of the coordinate system therefore the range is the set of uh non-negative real numbers or the set the zero to positive infinity okay uh for the graph of the function okay we define the graph of the function as the set of all points x y in the uh in the plane uh r square for which x y is an ordered pair in f or that's the locus of points okay all right so to easily trace the behavior of the uh to easily trace the graph of a function i identify some five basic steps the first one is to identify the dominant range and the properties of the function the second one is to choose the suitable values of x from the domain of a function and solve for its corresponding value of y third determine the behavior of x and y and fourth plot the points x and y and x y on the plane and five smoothly trace the curve okay let's try some examples say we want to sketch the graph of f of x equals 2x plus 3 we know that the function f of x equals to x plus 3 will give us a line okay and to trace or to sketch a line we might use the rise over run technique so here we will be getting the slope of the line and then we will be using the rise over run when we say rise that's the uh going up and going down and run going to the left or to the right okay so when the slope m is positive we can go up then right or down or then left and then when the slope m is negative we can go up then left then go down then right okay so we will be writing f of x equals to x plus three of the form y equals to x plus b and we notice that the slope is equal to 2 right m equals to 2 positive 2 that's positive 2 so 2 can be written of the form 2 over 1 so that's 2 unit rise and 1 unit run or negative two over negative one so i can go two units down or one unit left or two units up and one unit right okay so how do we start uh tracing this curve so we might this we might start at the intercept uh when x is zero so y equals three so we will be starting at uh three zero three okay and then so from here uh can i just go down and down and left okay that's the convenient here so i can i will go two units down and then one unit to the left okay so i might have a point somewhere there and then i since a line is determined by at least two points then i can now draw my line so this is the line uh f of x equals to x plus three and another way to to determine the domain of this function is uh you look for the values of x by which the graph will be passing through and the dominant this function is negative infinity to positive infinity and then the range is negative infinity to positive infinity or the set of real numbers and another way to sketch the graph of this function is by applying the intercept techniques so here we will be getting the x-intercept the x-intercept is the point on the x-axis by which the graph will pass through and here we will be setting y equals zero so when y equals 0 we solve x equal to negative 3 hub so we have uh an ordered pair negative 3 halves 0 to be the point on this line and then the y-intercept sorry that's negative 3 half 0 and then after which we will be looking for the y-intercept and the y-intercept is the point on the y-axis by which the graph passes through and here if we for the x-intercept we let y equal zero so for the y-intercept we will set x equals zero and then we look for y so when x is replaced by zero so 2 times 0 is 0 plus 3 so that is equal to 3 so y equals 3 so you have 0 3 and then we locate that here 0 3 and again since a line is determined by at least two points and then we can now easily trace the line uh two x plus three okay another one say we sketch the graph of f of x equals the square root of 25 minus x squared all right uh i will just be replacing f of x by y okay so we i now have y equals the square root of 25 minus x squared and this equation can be taught as y squared equals 25 minus x squared if you square both sides of this equation then you have y squared and then the square root will be avoided so you only have 25 minus x squared on the um right side all right and then manipulating this equation that lead us to x squared plus y squared equals 5 squared and recalling the properties of a conic section so x squared plus y squared equals 5 squared determines a circle with radius equal to five and whose center is located at zero zero again so this is x squared plus y squared equals five squared and this is a circle with center at the origin zero zero and that is just equal to five okay so this is five zero so from the center this is the center zero zero from the center you count five units one two three four five since uh our radius is equal to five and then from the center you go five units here five units to the left and then five units down and then you can easily trace the circle x squared plus y squared equals five squared so however what we are looking for is the function f of x equals 25 minus x squared okay so we notice that the domain of this function is the closed interval negative five to five okay and then the range is also negative five to five all right so however what we only need to look for is the function f of x equals the square root of 25 minus x squared so the idea is i'll be uh dividing this into two semi-circles we have this okay so this is now your y equals the square root of 25 minus x squared then again by the vertical line test we will notice that this function y equals 20 the square root of 25 minus x squared is really a function and then what about the semicircle below the uh x axis okay by the way the domain of this function is the interval negative five to five and then the range is from zero to five okay and then what about the semicircle below the x-axis so the equation would have been equal to y equals negative since we are now on the negative y axis so negative the square root of 25 minus x squared and again the domain of this function is the uh set the close interval negative five to five and then the range is from negative five to zero okay let's play along with this circle x squared plus y squared equals 25 okay again this is the circle and um if i divide this into two semicircles the one on the left side and the right side of the y axis so this semicircle is the equation x equals 25 the square root of 25 sorry minus y squared and this cannot be taught as a function since my vertical line test the vertical line passes through exactly two points on the curve and again the one written the one drawn on the uh left side left side of the y axis is the curve x equals negative the square root of 25 minus y squared and then the domain of this function is from negat the close interval negative 5 to 0 and then the range is the closed interval negative 5 2 5 right all right another one we sketch the graph f of x equals the square root of x squared minus four and again i will just replace f of x by y so i now have y equals the square root of x squared minus four and then again manipulating this equation squaring both sides i have now y squared equals x squared minus four and then eventually i can have x squared minus y squared equals positive four and again recalling the properties of the conics this is now a hyperbola with center at zero zero with my minor and major axis equal to two okay so this is uh negative to zero this is to zero and then we can now easily draw our hyperbola y squared equals x squared minus four or x squared minus y squared equals uh four okay and again what we are looking for is the function f of x equals the square root of x squared minus four again the hyperbola x squared minus y squared equals four cannot be taught as a function again by vertical line test it does not qualifies to be a function okay and then again what we only need to look for is the half of the hyperbola okay uh huh so this is now your f of x equals the square root of x squared minus four and then the domain of this function is the interval the open in closed interval negative infinity to negative two union the close open interval two to positive infinity and you will notice that the range of this function is from zero going to positive infinity okay another one say we sketch the graph of the function f of x equals 2 x squared over 25 minus x squared again this is a rational function of the form p of x over t of x so we should set this denominator 25 minus x squared not to be equal to zero so we have 25 minus x squared not equal to zero and then factoring this uh difference of two squares so you have five minus x and 5 plus x not equal to 0 so you obtain x equals 5 and x equals negative 5 so we can say that x cannot be 5 or negative 5 all right and then you will notice that the dominant this function is any number any real number except five or negative five so in interval notation we write the open interval negative infinity to negative 5 union negative 5 to 5 union 5 to positive infinity and we will verify that the range of this function is the interval negative infinity to negative 2 union close open interval zero to positive infinity okay so this is your x equals negative five so that is considered our vertical asymptote because uh x cannot be negative five another one is on the right side x equals five and this is another vertical asymptote okay since x cannot be five and then uh by the way uh another term for vertical asymptote is the line discontinuity or infinite discontinuity okay one more it's line discontinuity or infinite discontinuity okay so [Music] what about y equals negative 2 so this one is our horizontal asymptote since uh f of x cannot be negative 2 and how will thin uh and how do we obtain negative 2 uh huh so by getting the coefficients of x squared on the numerator and x squared on the denominator so that's 2 over negative 1 so you have y equals negative 2. so why do we need to do that one uh particular condition that we are going to do it is if the degree of the numerator is equal to the degree of the denominator then the horizontal asymptote are the equations of those two terms with equal degrees so in this case so this is 2x squared over negative x squared so the coefficients are 2 and negative 1 so 2 over negative 1 is negative 2. so the line y equals negative 2 is our horizontal asymptote and then we might uh trace this curve okay so this is the behavior of the curve and you will notice that the the curve to x squared over 25 minus x squared as x approaches a very large number to the left y gets closer and closer to negative two and similarly when x gets closer and closer to negative 5 y approaches negative infinity on the other side as x gets larger and larger going to positive infinity y gets smaller and smaller nearly approaching uh negative two and similarly when x gets closer and closer to negative five your y decreases going to negative infinity and other part of the curve is this uh parabola okay uh you will notice that when x is zero when x is zero obviously y is also equal to zero and this is the lowest point of this uh part of the uh curve all right uh sketch the graph of the function f of x equals 1 over x squared minus 16. again taking the idea that the denominator should not be equal to 0 so we know that x cannot be negative four or positive or okay so this is your x equals negative four your x equals four and then these two are the vertical asymptotes or the infinite discontinuity and then the line y equals zero is your horizontal asymptote where do we get this y equals zero you will notice now that the numerator is a constant one okay so this is one so therefore uh the degree of the numerator the degree of the numerator uh which is zero is less than the degree of the denominator then the y the x axis is horizontal asymptote again earlier you notice that the the on our previous example that's 2x squared over 25 minus x squared the degree of the numerator uh by the way the degree is the highest power or the highest exponent on the numerator and denominator so in our previous example 2x squared over 25 2x squared over 25 minus x squared you notice that the degree of the numerator 2 is equal to the degree of the denominator that's the reason why the horizontal asymptote is y equals negative 2. our next example uh the degree of the numerator is 0 and the degree of the denominator is 2 and the degree of the numerator is less than the degree of the denominator so therefore the x-axis is a horizontal asymptote okay and then we might trace the curve okay and the domain of this function is the set of real number except plus or minus four and then the range is the open and close interval negative infinity to negative 1 over 16 union the open interval 0 to positive infinity and that is confirmed by the graph since there are part of the curve graph trace on the negative x a negative y so you start at negative infinity going to this point and this is your negative 1 over 16. how do we get the negative 1 over 16 obviously when x is 0 when this is 0 so you have 1 over negative 16. so 0 negative 1 over 16. uh-huh i think this would be the last we sketch the the graph of the function f of x equals x squared minus 9 over x plus three okay again this is a rational function so we might say that x plus three cannot uh be equal to zero okay so we say that x cannot be negative three okay because when x is negative three the denominator will become zero okay however you notice that x squared minus nine all over x plus three is not in lowest term okay why because we can factor out x squared minus nine of this form x plus three x minus three and then all over x plus three so you can now cancel this out so we only left with x minus three so f of x can now be thought as x minus three so this is again a line whose graph okay this is the graph of the line y equals x minus three however we identified earlier that x cannot be negative 3 so x cannot be uh sorry negative 3 cannot be used to replace x here so we might have a hole there okay and this one is called the point discontinuity so when x is negative 3 so y is negative 6 so from here you replace x by negative 3 so y equals negative 3 minus 3 and that leads you to y equals negative 6. since x cannot be negative 3 therefore y equals negative 6 that is the reason why we have a hole here okay so again this is what we call the point discontinuity and again the domain of this function is the set of real numbers except negative three and then the range is the set of real number except negative six okay well i have mentioned earlier the asymptotes and this is a basic definition of an asymptotes how do you determine whether uh a line is a vertical asymptote or a horizontal asymptote uh suppose that f of x equals p of x over q of x is a rational function in those terms meaning the numerator and denominator do not have any common factors and a is is some real number where k of x is not equal to zero then the vertical line x equals a is a vertical asymptote of the graph of f if as x approaches a then the absolute value of f of x approaches infinity okay and the horizontal line y equals a is a horizontal asymptote of the graph of f if as the absolute value of x approaches positive infinity then f of x approaches a all right suppose that the rational function f of x equals p of x over t of x and where p of x is written of this form a sub n x raised to n plus a sub n minus 1 x raised to n minus 1 running until a sub 1 x plus a sub 0. and q of x is written of the form b sub m x raised to m plus b sub n minus 1 x raised to m minus 1 running until b sub 1 x plus b sub 0 and the where key of x is not equal to zero and that p of x over q of x is in lowest terms and here are the qualification of to become an asymptote okay if p of a equals zero then x equals a is a vertical asymptote okay next if n is less than m meaning the degree of the numerator is less than the degree of the denominator then the x axis is horizontal asymptote next if n equals m then the horizontal asymptote is the line y equals a sub n over b sub n meaning we just get the coefficient of x to the n and x to the m so that's a sub n over b sub m and if n is greater than m then the graph of f has no horizontal asymptote meaning if the degree of the numerator is larger than the degree of the denominator then the graph of f has no horizontal asymptote and that is observed from our previous examples all right another one say f of x equals x squared plus 5x plus 10 over 2x plus 6. we will be determining whether this function has a horizontal asymptote or vertical asymptote okay so we let p of x equals x squared plus 5x plus 10 and t of x equals 2x plus 6 and we notice that the degree of p of x n equals 2 and that the degree of q of x m equals 1 so that's the highest power here x squared plus 5 x plus 10 the highest exponent is 2 so that's the degree and then for 2x plus 6 the highest power is equal to 1 okay so since uh the degree of the numerator n is larger than the degree of the denominator 2 is greater than 1 therefore this function has no horizontal asymptote okay and for the vertical asymptote we will be getting 2x plus 6 and we equate it to 0 so we'll say x equals negative 3. so the line x equals negative 3 is a vertical asymptote now say f of x equals p of x over t of x can be can be reduced of the form g of x plus r of x over q of x this g of x plus r of x q of x can be obtained if you divide p of x over q of x when are you going to divide uh the fraction or or when are you going to divide p of x by k of x that is when the degree of p of x is greater than or equal to the degree of q of x so when you divide p of x by q of x you have a quotient plus remainder over divisor so q of x is the quotient and r of x is the remainder okay and uh if this g of x is a line of the form m x plus b okay if this g of x can be written of this form m x plus b then the line y equals m x plus b is an oblique asymptote so you now have vertical asymptote and oblique asymptote and horizontal asymptote okay so obviously this function f of x equals x squared plus 5 x plus 10 over 2x plus 6 has a vertical asymptote and that is the line x equals negative three however it has no horizontal absence so what are we going to do now is to identify whether this function has an oblique asymptote okay so what are we going to do is we will be dividing this x squared plus 5x plus 10 by 2x plus 6. by long method of division okay i used long method of division here so we have when x squared plus 5x plus 10 is divided by 2x plus 6 the quotient is 1 half x plus 1 and then the remainder is 4. okay so we now have f of x can be written as one half x plus one plus four all over two x plus six and then since this uh one half x plus one is a line of the form y equals m x plus b therefore the line y equals one half x one of x plus one is an oblique asymptote okay so we recall that the vertical asymptote of this uh function is the line x equals negative three and then we trace this curve all right so this is the line x equals negative three and then obviously again there is no horizontal asymptote but we have an oblique asymptote of the form y equals one half x plus one and this is the line y equals one of x plus one okay so we can now easily trace the the graph uh x squared plus 5x plus 10 over 2x plus 6 and obviously it will not pass through this lines x equals negative 3 and y equals 1 of x plus 1. and uh we set we assume the values of x say when x is zero the question now is will this function passes through the y axis so when x is zero zero zero and x is zero so you have ten over six ten over six that's approximately five uh five thirds so that's one point sixty seven so we lit we locate that zero 0 1.67 and we uh find it there right that's 1.67 and then we can easily trace the curve okay and you will notice that since this is an asymptote the the behavior of this curve will just get closer and closer to this line but will never pass through and then it will also get closer and closer to x equals negative and will never pass through and this curve will have a reflection over this side of the cartesian coordinate system okay and that completes the graph f of x equals x squared plus 5x plus 10 all over 2x plus 6. and that's it that's it for now and uh on our next video just keep on tuning into my channel i'll be presenting to you how to uh perform operations and functions and the different types of functions and that's it for now and uh i'll see you next on my previous on on my next videos goodbye