Question

If f(x) = ∣(x2 − 6)(x2 + 2)∣, how many numbers in the interval 1 ≤ x ≤ 2 satisfy the conclusion of the mean value theorem?

Answer 1

\[f'(c)=\frac{f(2)-f(1)}{2-1}\]
is what you need to solve

Answer 2

I know that but I understand that you need the calculate the derivative of the equation and that is what i'm confused about and also how to solve to find the values of c.

Answer 3

\[f(x)=|(x^2+2)(x^2-6)| \\ x^2+2 \ge 0 \text{ for all real x } \\ f(x)=(x^2+2)|x^2-6| \\ \text{ we only care what happens with f when x is between 1 and 2 } \\ x^2-6=0 \text{ when } x=\pm \sqrt{6}\]
so our numbers occur between the two zeros (or are interval is a subinterval of (-sqrt(6),sqrt(6)) so we need to find out what is going on between there

Answer 4

like is the function positive or negative on that interval

Answer 5

i flipe the y values that corresponded to the x-values between -sqrt(6) and sqrt(6)