Question

f(x)=x(x^2-x-2) Determine whether the Mean Value Theorem can be applied to f on the closed interval [a,b]. If the Mean Value Theorem can be applied, find all values of c in the open interval (a,b) such that f'(c)=f(b)-f(a)/b-a

Answer 1

Check if the hypothesis of the MVT is satisfied by f. Then just solve the equation for c

Answer 2

is it
\[f(x)=x(x^2-x-2)=x^3-x^2-2x\]?
all polynomials are continuous for all real numbers, so the mvt applies to any polynomial over any interval

Answer 3

do you have specific values for \(a\) and \(b\) ?

Answer 4

The interval is [-1,1]

Answer 5

is it the function i wrote?

Answer 6

Yes, that is the function

Answer 7

the cubic function
ok lets take it step by step

Answer 8

Yes, please!

Answer 9

you need \(f(1)=1^3-1^2-2\times 1=-2\)

Answer 10

I wish I understood that. This is what I've done:
f'(x)= 3x^2-2x-2
f(-1)=3(-1)-2(-1)-2=3
f(1)=3(1)^2-2(1)-2=-1

Answer 11

Oh okay, I see what you did

Answer 12

you also need \(f(-1)=(-1)^3-(-1)^2-2\times (-1)=0\)

Answer 13

it is the function evaluated at the endpoint of the derivative, not the derivative

Answer 14

so \(f(b)-f(a)=f(1)-f(-1)=-2-0=-2\)

Answer 15

and
\[\frac{f(b)-f(a)}{b-a}=\frac{f(1)-f(-1)}{1-(-1)}=\frac{-2}{2}=-1\]

Answer 16

your final job is to take the derivative, set it equal to \(-1\) and solve the resulting (quadratic) equation
you good from there?

Answer 17

So 3x^2-2x-2=-1 will I solve this using quadratic equation? (so sorry you're having to simplify this so much!)

Answer 18

yes

Answer 19

the derivative is \(f'(x)=3x^2-2x-2\) and you want to know for what number in the interval \([-1,1]\) that it is equal to \(-1\) so set
\[3x^2-2x-2=-1\] and solve for \(x\) via
\[3x^2-2x-1=0\] etc

Answer 20

by some miracle this problem was cooked up to factor, so you can use
\[(3x+1)(x-1)=0\] to solve it

Answer 21

Thank you so, SO much!