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Q. 58

Expert-verified
Found in: Page 249

### Calculus

Book edition 1st
Author(s) Peter Kohn, Laura Taalman
Pages 1155 pages
ISBN 9781429241861

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

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

See the step by step solution

## Step 1. Given information

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

## Step 2. Proving Mean Value Theorem.

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

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

localid="1648379095216"

By the Mean Value Theorem, there must exist at least one point $c\in \left(0,3\right)$ with role="math" localid="1648378574205" $f\text{'}\left(c\right)=\frac{8}{3}$

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

role="math" localid="1648379138019" ${2}^{c}\mathrm{ln}2=\frac{7}{3}$

role="math" localid="1648379195217" ${2}^{c}=\frac{7}{3×0.693}\phantom{\rule{0ex}{0ex}}{2}^{c}=3.36$

Therefore,

$c=1.75$