Q. 55 - Calculus

Question

Determine whether or not each function f in Exercises 53–60 satisfies the hypotheses of the Mean Value Theorem on the given interval [a, b]. For those that do, use derivatives and algebra to find the exact values of all c ∈ (a, b) that satisfy the conclusion of the Mean Value Theorem. 

$f\left(x\right)=-x^3+3x^2-7,[a,b]=[-2,3]$

Step-by-Step Solution

Answer

The function satisfies the mean value theorem and $c=2.527,-0.527$

Step 1. Given information

$f\left(x\right)=-x^3+3x^2-7$

Step 2. Proving Mean Value Theorem

$\begin{array}{r}{f}^{\mathrm{\prime }}\left(c\right)=\frac{f\left(3\right)-f(-2)}{3-(-2)}\\ =\frac{-3^3+3\cdot 3^2-7-\left[-(-2{)}^3+3\cdot (-2{)}^2-7\right]}{5}\\ =\frac{-7-8-12+7}{5}\\ =-4\end{array}$

Now,

$\begin{array}{r}{f}^{\mathrm{\prime }}\left(c\right)=-4\\ -3c^2+3\cdot 2c=-4\\ 3c^2-6c-4=0\end{array}$

$\begin{array}{r}c=\frac{-(-6)\pm \sqrt{(-6{)}^2-4\cdot 3\cdot (-4)}}{2\cdot 3}\\ =\frac{6\pm \sqrt{84}}{6}\end{array}$

$\begin{array}{r}=\frac{6\pm 9.165}{6}\\ =2.527,-0.527\end{array}$