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Q. 59

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 $f$ 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 $c\in (a,b)$ that satisfy the conclusion of the Mean Value Theorem.

$f\left(x\right)=\mathrm{sin}\left(x\right),[a,b]=[0,\frac{\mathrm{\pi}}{2}]$.

The function $f\left(x\right)=\mathrm{sin}\left(x\right)$ satisfies the Mean Value Theorem and the value is, $c={\mathrm{cos}}^{-1}\left(\frac{2}{\mathrm{\pi}}\right)$.

The given function is,

$f\left(x\right)=\mathrm{sin}\left(x\right),[a,b]=[0,\frac{\mathrm{\pi}}{2}]$.

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

The slope of the line from $(0,f(0\left)\right)$ to $(\frac{\mathrm{\pi}}{2},f(\frac{\mathrm{\pi}}{2}\left)\right)$ is:

$\frac{f\left(\frac{\mathrm{\pi}}{2}\right)-f\left(0\right)}{\frac{\mathrm{\pi}}{2}-0}=\frac{\mathrm{sin}\left(\frac{\mathrm{\pi}}{2}\right)-\mathrm{sin}\left(0\right)}{\frac{\mathrm{\pi}}{2}-0}\phantom{\rule{0ex}{0ex}}=\frac{1-0}{\frac{\mathrm{\pi}}{2}-0}\phantom{\rule{0ex}{0ex}}=\frac{1}{\frac{\mathrm{\pi}}{2}}\phantom{\rule{0ex}{0ex}}=\frac{2}{\mathrm{\pi}}$

By the Mean Value Theorem, there must exist at least one point $c\in (0,\frac{\mathrm{\pi}}{2})$ with ${f}^{\text{'}}\left(c\right)=\frac{2}{\mathrm{\pi}}$

We have to find the value of $c$ with ${f}^{\text{'}}\left(c\right)=\frac{2}{\mathrm{\pi}}$ we solve it:

$\mathrm{cos}\left(c\right)=\frac{2}{\mathrm{\pi}}\phantom{\rule{0ex}{0ex}}\Rightarrow c={\mathrm{cos}}^{-1}\left(\frac{2}{\mathrm{\pi}}\right)$

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