Determine if the Mean Value Theorem applies, then find "c" such that f'(c)=(f(b)-f(a))/(b-a)

Question

anonymous · Answer

f(x)=x^3-x^2-2x for [-1,1]

anonymous · Answer

why would the answer be c=-1/3 and not 1 @TweT226

anonymous · Answer

you have to take the derivative

anonymous · Answer

$$f'(c)=\frac{ f(b)-f(a) }{ b-a }$$
$$3c^2-2c-2=\frac{ -2-0 }{ 1- (-1)}$$

anonymous · Answer

$$3c^2-2c-2=-1$$
$$3c^2-2c-1=0$$
$$(3c+1)(c-1)=0 $$
$$c=\frac{ -1 }{ 3 }, 1$$

anonymous · Answer

so how come my teacher chose $$\frac{ -1 }{ 3}$$ over c=1?

anonymous · Answer

I think it's because it's right on the endpoints

anonymous · Answer

what do u mean

anonymous · Answer

i mean interval [-1, 1]

anonymous · Answer

the theorem states that the equation is differentiable between (a,b) not [a,b] the  domain is [=1,1] in the problem, so when you take the derivative, c cannot equal a or b.

anonymous · Answer

oh u mean it's in between

anonymous · Answer

oh ok gotcha ;)

anonymous · Answer

-1*

anonymous · Answer

yeah

anonymous · Answer

@yun2thejae can u help me with another problem pls? ^^

anonymous · Answer

sure, if you would like.