this is lesson 4.2 in which we discuss really just an introduction to instantaneous rates of change and the main goal of this lesson is to practice and calculate difference quotients this is a topic that students struggle with in calculus and so it's great to practice in our ramp up so um the question what is the difference quotient well last lesson we discussed average rate of change on an interval and the difference quotient is an average rate of change but it's on this interval from x to x plus h and i've drawn a picture here i have a function f x in blue and here is a x and here is some x plus h so the points on the graph are here and here and we know the average rate of change on that interval well it is the slope of what's called the secant line here is this line in black but let's write what the slope is so the average rate of change of a function from x to x plus h it would be f of x plus h minus f of x okay that's the change in y and then divided by the change in x which is x plus h minus x now we can simplify this sum because you notice in the denominator i have x minus x which is zero and so we just get this f of x plus h minus f of x and then all divided by h this is what a difference quotient is and maybe i'll do this in black this is the slope of this line this line has slope m where this is the same m i verbally said secant line but let me label it this black line is our secant line from x to x plus h it's kind of small but this is how they are labeled and the slope of this secant line is this difference quotient right here f of x plus h minus f of x divided by h and this is really our goal to practice calculating and simplifying these using algebra in this lesson well here's a few examples the first one we have f of x is minus 10x plus seven find the average rate of change from x to x plus h well we know we just discussed this this is f of x plus h minus f of x and then divided by h well the strategy for calculating and simplifying some of these difference quotients typically i will work with just the numerator first so let's come down here and calculate f of x plus h minus f of x here we have minus 10 times x plus h plus seven and then subtract off f of x so we have minus 10 x plus seven and don't forget here this minus sign it distributes across the parentheses so all together let's multiply out we have minus 10x minus 10h there's a plus seven and then we have a plus 10x minus seven um one more step here before this is simplified we have a maybe i should underline we have a minus 10x and a plus 10x those add to zero plus seven and minus seven and those add to zero so all together this f of x plus h minus f of x which is the change in y we just get minus ten h okay now we're ready to jump into the difference quotient the quotient part of this because we have minus 10h this is our numerator and then divided by h this is minus 10. now does this make sense well yes this is a straight line with slope negative 10 it's at the form mx plus b and so we know the slope of a line is its rate of change and so it definitely makes sense to see the slope here as our difference quotient the average rate of change from x to x plus h okay fantastic let's move on we have a quadratic this time we want the average rate of change from 2 to 2 plus h so here x is 2. all right what we want just as i did above we want f of 2 plus h minus f of 2 and then divided by h so very similarly to the example we just did i'm first going to work with the numerator and simplify so here f of 2 plus h minus f of 2 f of 2 plus h we evaluate our function which is here at 2 plus h so everywhere we see an x we put in 2 plus h just like this now subtract off f of two so if we subtract off f of two is three times 2 squared minus 2. well we have some work to do here in particular we have to multiply all of this out this is not 2 squared plus h squared we discussed this in our exponents lesson so this we need to remember is 2 plus h times 2 plus h maybe for the moment i will leave the 3 out in front of this we get three so my first is four you notice the inner and the outer are both 2h so i have 2 times 2h which is 4h then the last is h squared and then i subtract off two subtract off h now this part maybe i'll write above and pink what we have this is four times 3 which is 12 minus 2 so this part is 10. we have 12 plus 12 h plus 3 h squared and then minus h minus 12. one more step and i'll be ready to move up to the blue and calculate the difference quotient well we have 12 minus 12 adds to zero and that's about it so i'm left with 11 h plus 3 h squared so let's move back up and we will calculate the difference quotient we have found the numerator maybe as i write this because you see i have an h in the denominator so i'm going to factor out an h from this quantity we have h and then times 11 plus 3 h and now the last step we have 11 plus 3h so this is our simplified average rate of change of this function from 2 to 2 plus h and we use a difference quotient here next example we want to find the average rate of change of 2 over x from 3 to 3 plus h so once again let's write what this is we want f of 3 plus h minus f of 3 divided by h now just like in the last two examples i start with the numerator and simplify as much as i can so my numerator is um f of 3 plus h minus f of three we evaluate my function at three plus h so i have two divided by three plus h then subtract off evaluate at 3 we have 2 over 3. now the question is how do you simplify this because i have a sum of two fractions or a difference really of two fractions so what i would do here and this is the strategy you can use when you see these types of problems in calculus difference quotients where you have a fraction is i would get a common denominator and then we can add so the common denominator here is the product of the two denominators so what i'm going to do maybe i will spread some stuff out and i can do this in pink i will multiply this one by 3 plus h divided by 3 plus h this is 1 and i will multiply this one by three over three this is also one so i'm not changing anything i'm just in the process of getting a common denominator and we can see my common denominator here it's three times three plus h now my numerator well maybe let me add the parentheses here so i don't forget that they are there my numerator we see is 6 and then minus two times three plus h so i can take a few steps here we have six minus six minus two h divided by this common denominator three and three plus eight one more step and then i will jump into the difference quotient in blue you notice six minus six is zero so i just get negative two h okay so this is the numerator of my difference quotient and i will move it down just like this now what happens when you divide this by h well let's copy and paste and we can write where the h really is the h is in the denominator so this will be here and then we get minus 2 divided by 3 times 3 plus h this is our final answer just like this now comes the question what is an instantaneous rate of change and this is really fundamental to calculus instantaneous rates of change you will study this so much in your calculus courses but to get an idea for this think about driving a car down a road and getting a speeding ticket let's hope you don't by getting a speedy ticket for 70 miles per hour this is not averaging over some interval this is in a very moment you are going 70 miles per hour and this is our instantaneous rate of change it says in an instant or instantaneously at one time value or x value as a function of x what is the rate of change and this is what we want to discuss now how could we calculate such a thing well the average rate of change from x to x plus h is this difference quotient f of x plus h minus f of x over h and we've practiced a few of these in this video we will practice a couple more but then you let h approach zero now you notice in this you cannot just evaluate h equals zero h equals zero we would get this f of x minus f of x over zero this is zero over zero this is what's called an indeterminate form so that is not the way to calculate this just evaluate at zero but if you let h approach zero this is how we measure the instantaneous rate of change um here is a schematic where i have x x plus h now let h go to zero here is another picture and then here they're very close together but for this video let us look in desmos so here i have this exact same thing but graphed in desmos where i can move it so here okay the slope of this black line is the difference quotient is the average rate of change from x to x plus h of this function and if you just think about slopes we just let h approach zero so you notice the slope of this line is changing but it's approaching something and it's approaching the instantaneous rate of change at that x value and so this is how we define this okay now let's go back so let me add in some words as to how we will calculate this so we want to simplify this until the expression is defined at h equals zero and then we substitute h equals zero so you notice this is not defined at h equals zero but quite often we can simplify until it is and then we just evaluate at age equals zero and we have the instantaneous rate of change for example here this is defined at h equals zero to get the instantaneous rate of change at x equals two we just substitute h equals zero we get eleven okay really what's happening is something called a limit which you'll learn about in calculus but for the purpose of simplifying a difference quotient and evaluating at h equals zero this is the discussion that we want to have now really this is just a remark here before we do a few more examples given a function we define what's called a derivative of the function and it's denoted f prime of x and it satisfies that f prime of a is the instantaneous rate of change at x equals a so this is what you will start off in calculus one and study derivatives a lot of derivatives in calculus 1. but this function as you will see and learn more about is really giving us instantaneous rate of change so let's do a few examples here we have a student there's a book in the air and the book's height as a function of time is given by this p of t it's quadratic it's feet empty seconds we want to find an average velocity from t equals zero to t equals one and then an instantaneous velocity at t equals one well this first one is really a nice review of the previous lesson this would just be p of one minus p of zero and then divided by one minus zero this is average rate of change on an interval p of one is minus sixteen plus fifty plus five which is 39 and p of 0 is 5. evaluate at 0. so here we get 39 minus 5 over 1. so we have 34 and you can think about units here change in p over change in t and p is in feet t is in seconds so this would be feet per second maybe i can add this notation we talked about in the last lesson change in p divided by change in t okay wonderful now here comes what's new for this lesson the instantaneous velocity at t equals one what we want we want p of one plus h minus p of one over h and then we let h approach zero this is the instantaneous rate of change well let's evaluate p of the numerator just like we were practicing p of one plus h minus p of one p of one plus h is minus sixteen one plus h squared plus 50 one plus h and then plus five and then we subtract off p of one but you see we've already calculated that above p of one is thirty nine now just like our previous example where we had a quadratic this is 1 plus h times 1 plus h so let's multiply this out we have minus 16 we have 1 plus 2h plus h squared and then we have 50 plus 58 and then this is a negative 34. we can take a few more steps before we divide by h so i have negative 16 minus 32 h minus 16 h squared then 50 minus 34 is plus 16 and i have a plus 50 h here's where we realize things are going to work out wonderfully for us because we have minus 16 and plus 16 we're left with 18 h minus 16 h squared and this is h times 18 minus 16 h well now we can calculate this difference quotient p of 1 plus h minus p of 1 divided by h you see we just take this expression and divide by h and so we have 18 minus 16 h period so as h approaches 0 you see we can just evaluate we get positive 18 and this would be also the same units feet per second this is our answer this is our instantaneous rate of change at t equals 1 which is our instantaneous velocity we have one more example we want the instantaneous rate of change of one over two x plus three at x equals one so we calculate g of one plus h minus g of one over h and then we let h approach zero which means we have to simplify this difference quotient such that we can evaluate at h equals zero so let's get started with my numerator do you have one plus h minus g of one well we have one divided by we have two one plus h plus three now subtract off g of one which is one over two plus three now before i get a common denominator i wanna take one step with both of these you see this we have two plus two h plus three all together we have two h plus five so this is one divided by two h plus five and subtract off one divided by five here's where we're ready for a common denominator just like we did in a previous example that involved a fraction so this one i'll multiply by two h plus five over two h plus five and this first term i multiply by five over five my common denominator is here it's two h plus five times five now i have a five then minus we have two h plus five you see i'm in a really wonderful shape because this is going to be five minus five adds to zero so my numerator here is just negative two h we have negative 2h and then my denominator is this product which i'll leave just like this 2h plus 5 times 5. well this is h times something it's h times all of this so i'm ready to divide by h here we can have this difference quotient simplified such that it's defined at h equals zero and the way to do this is we just divide all of this by h so now my numerator is just negative two and my denominator is here it's two h plus five times 5. okay as h approaches 0 what do we get with this difference quotient just evaluate at zero we get 2 minus 2 part of v over 5 times 5 which is negative 2 over 25 and this is our instantaneous rate of change here we don't have units given in the problem so i won't write any units but negative 2 over 25 this is our instantaneous rate of change of this function at x equals 1. so i will underline this this is our last example and you can move on work on the exercises next thank you so much students