Question

Determine whether or not each function f in Exercises 41–48 satisfies the hypotheses of Rolle’s 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 Rolle’s Theorem.

$f\left(x\right)=e^x\left(x^2-2x\right),\left[a,b\right]=\left[0,2\right]$

Short answer

The function $f\left(x\right)=e^x\left(x^2-2x\right)$ satisfies the hypotheses of Rolle's theorem. The exact values of c that satisfies conclusion of Rolle's theorem is $c=\sqrt{2}$.

Step-by-step solution

Step 1. Given information.

Consider the given function $f\left(x\right)=e^x\left(x^2-2x\right),\left[a,b\right]=\left[0,2\right]$.

Step 2. Satisfy hypotheses of Rolle's theorem.

The given function is continuous on $\left[0,2\right]$and differentiable on $\left(0,2\right)$.

Now, find $f\left(0\right)$and $f\left(2\right)$.

$\begin{array}{rcl}f\left(0\right)& =& e^0\left(0^2-2\left(0\right)\right)\\ & =& 0\\ f\left(2\right)& =& e^2\left(4-4\right)\\ & =& 0\end{array}$

So, $f\left(0\right)=f\left(2\right)$.

Thus, Rolle's theorem applies on$\left[0,2\right]$then there must exist some value c such that $f\text{'}\left(c\right)=0$.

Step 3. Find the exact values of c.

The function is $f\left(c\right)=e^c\left(c^2-2c\right)$.

Differentiate the function with respect to c.

$f\text{'}\left(c\right)=e^x{x}^2-2e^x$

Solve $f\text{'}\left(c\right)=0$.

$$\begin{gathered} e^c{c}^2-2e^c=0 \\ c=\sqrt{2} \end{gathered}$$