f(x) is increasing if f'(x) is positive
so (x−8)^7(x−1)^4(x+19)^8 > 0
we know that (x−1)^4(x+19)^8 is always positive since we have even exponents.
So all we have to do is look at
(x-8)^7 > 0
if x > 8 then (x-8)^7 > 0
if x < 8 it is negative
so the function is increasing for
x > 8
However, when x = 1 or -19, f'(x) = 0, so the function would not be increasing, but those two values would be excluded anyway by x > 8
so it is increasing for all real values of x such that
x > 8
Suppose the derivative of a function f is f′(x)=(x−8)^7(x−1)^4(x+19)^8.
Then the function f is increasing on the interval what?
1 answer