Given the function #f(x)=-x^3+4x^2-3#, how do you determine whether f satisfies the hypotheses of the Mean Value Theorem on the interval [0,4] and find the c?

Question

Cameron Alexander · Accepted Answer

The values of #c# are #={0,8/3}#

#f(x)# is a polynomial function.

So it is continuous on the interval #[0,4]#and differentiable on the interval #]0,4[#, therefore we can apply the mean value theorem which states that

there is #c in [0,4]# such that

#f'(c)=(f(4)-f(0))/(4-0)#

#f(x)=-x^3+4x^2-3#

#f(0)=-0+0-3=-3#

#f(4)=-64+64-3=-3#

Therefore,

#f'(c)=(f(4)-f(0))/(4-0)=(-3-(-3))/(4)=0#

Also,

#f'(x)=-3x^2+8x#

#f'(c)=-3c^2+8c=0#

Solving for #c#

#c(-3c+8)=0#

So,

#c=0#

and

#c=8/3#

Both values of #c in [0,4]#

Luke Alvarez · Answer

undefined To determine if the function $ f(x) = -x^3 + 4x^2 - 3 $ satisfies the hypotheses of the Mean Value Theorem on the interval $[0, 4]$, we first check if the function is continuous on the closed interval $[0, 4]$ and differentiable on the open interval $(0, 4)$.

The function is a polynomial, so it is continuous and differentiable everywhere. Therefore, it satisfies the hypotheses of the Mean Value Theorem.

To find the value of $ c $ guaranteed by the Mean Value Theorem, we calculate the derivative of $ f(x) $ and then find the average rate of change of $ f(x) $ over the interval $[0, 4]$. This average rate of change is equal to the slope of the secant line connecting the endpoints of the interval.

First, find the derivative of $ f(x) $:
$$ f'(x) = -3x^2 + 8x $$

Next, find the average rate of change of $ f(x) $ over the interval $[0, 4]$:
$$ \frac{f(4) - f(0)}{4 - 0} = \frac{f(4) - f(0)}{4} $$

$$ = \frac{-4^3 + 4(4)^2 - 3 - (-0^3 + 4(0)^2 - 3)}{4} $$

$$ = \frac{-64 + 64 - 3 - (-3)}{4} $$

$$ = \frac{-3}{4} $$

This is the slope of the secant line. Now, we find the value of $ c $ such that the derivative of $ f(x) $ evaluated at $ c $ is equal to this average rate of change.

$$ f'(c) = -3c^2 + 8c = \frac{-3}{4} $$

Solve this equation for $ c $ to find the value guaranteed by the Mean Value Theorem.