Short Answer

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

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

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1. Given information

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

Step 2. Proving Mean Value Theorem.

The function $f (x) = 2^{x}$ 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 $(3, f (3))$ is:

localid="1648379095216" $\frac{f (3) - f (0)}{3 - 0} = \frac{2^{3} - 2^{0}}{3 - 0} = \frac{8 - 1}{3 - 0} = \frac{7}{3}$

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

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

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

role="math" localid="1648379195217" $2^{c} = \frac{7}{3 \times 0.693} 2^{c} = 3.36$

Therefore,

$c = 1.75$

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