Question

If f(x) = |(x2 - 10)(x2 + 1)|, how many numbers in the interval 0 ≤ x ≤ 2.5 satisfy the conclusion of the mean value theorem?

Answer 1

I assume they are asking you how many points the derivative is equal to the average rate of change.

Answer 2

First find: \[
\frac{f(2.5)-f(0)}{2.5-0}
\]

Answer 3

Then find all \(x^*\) such that: \[
f'(x^*) = \frac{f(2.5)-f(0)}{2.5-0}
\]

Answer 4

Remember that: \[
\frac{d}{dx}|x| = \frac{|x|}{x}
\]By the chain rule: \[
|f(x)|' = \frac{|f(x)|f'(x)}{f(x)}
\]

Answer 5

so f '(c) = 1 ?

Answer 6

Does that say
f(x) = |(x^2 - 10)(x^2 + 1)|

Answer 7

yes

Answer 8

did you follow wio's suggestion and calculate the average rate of change?

Answer 9

yes and I got 1 for that

Answer 10

you give try to get rid the absolute value before differentiating f
x^2+1 is always positive so
f(x)=(x^2+1)|x^2-10|
we are only conccerned what happens on x=0 to x=2.5
y=x^2-10 is a parabola that is faced up with x-intercepts -sqrt(10) and sqrt(10)
with a y-intercept of -10
so between x=0 to x=2.5 the parabola is underneath the x-axis
so there |x^2-10|=-(x^2-10)

so I would look at differentiating f(x)=-(x^2+1) (x^2-10)
but either way