Aug 8, 2024
f(x) = -7x + 5
, evaluating f(2)
involves replacing x
with 2
to get -7*2 + 5 = -9
.f(2)
vs. y = -7x + 5
– Function notation is preferred for clarity and descriptiveness.f(x) = -7x + 5
and g(x) = 3/2x - 1
are different functions despite using x
as the variable.h(x) = -2x^2 + x - 1
, evaluating h(-1)
involves replacing x
with -1
.sqrt(x^2 - 3x)
at x = 1
results in sqrt(1 - 3) = sqrt(-2)
, which is not a real number.g(x) = (2x + 1)/(x - 5)
, x = 5
is outside the domain because it makes the denominator zero.(f(x + h) - f(x)) / h
.f(x + h)
by replacing x
with x + h
.f(x + h)
and f(x)
in the difference quotient formula.h
if possible.f(x) = x^2 + 1
, f(x + h) = (x + h)^2 + 1
simplifies to x^2 + 2xh + h^2 + 1
. The difference quotient then simplifies to 2x + h
.5
, 0
, or -1
to understand the output.2x
or x + 1
as inputs to test the function's flexibility.