Q. 58 - 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.

$f\left(x\right) = 2^x, [a, b] = [0, 3]$

Step-by-Step Solution

Answer

The function satisfies the Mean value theorem and the value of c is 1.75.

Step 1. Given information

$f\left(x\right) = 2^x, [a, b] = [0, 3]$

Step 2. Proving Mean Value Theorem.

The function $f\left(x\right)=2^x$ is continuous and differentiable on $[0, 3]$. The Mean Value Theorem applies to this function on the interval $[0,3]$.

The slope of the line from (0, f(0)) to $(3, f(3\left)\right)$ is:

$$\begin{gathered} \frac{f\left({\displaystyle 3}\right)-f\left(0\right)}{{\displaystyle 3-0}}=\frac{2^3-2^0}{{\displaystyle 3-0}} \\ =\frac{8-1}{{\displaystyle 3-0}} \\ =\frac{7}{{\displaystyle 3}} \end{gathered}$$

By the Mean Value Theorem, there must exist at least one point $c\in (0,3)$ with  $f\text{'}\left(c\right)=\frac{8}{3}$

We have to find the value of c with $f\text{'}\left(c\right)=\frac{8}{3}$ we solve it:

$2^c\ln 2=\frac{7}{3}$

$$\begin{gathered} 2^c=\frac{7}{3\times 0.693} \\ {2}^c=3.36 \end{gathered}$$

Therefore,

$c=1.75$