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Q. 58

Expert-verifiedFound in: Page 249

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.

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

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

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

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:

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

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

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" ${2}^{c}=\frac{7}{3\times 0.693}\phantom{\rule{0ex}{0ex}}{2}^{c}=3.36$

Therefore,

$c=1.75$

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