Verify mean value theorem for the function: f(x)= - 2x in [0, ]. - Vidyadip

Question

Verify mean value theorem for the function:
$\text{f(x)}=\sin\text{x}-\sin2\text{x in }[0,\pi].$

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Answer

We have, $\text{f(x)}=\sin\text{x}-\sin2\text{x in }[0,\pi]$

Since, we know that sine functions are continuous functions hence

$\text{f(x)}=\sin\text{x}-\sin2\text{x}$ is a continuous function in $[0,\pi].$

$\text{f}'(\text{x})=\cos\text{x}-\cos2\text{x}.2=\cos\text{x}-2\cos2\text{x},$ which exists in $(0,\pi)$

So, f(x) is differentiable in $(0,\pi).$ Continuous of mean value theorem are satisfied.

Hence, $\exists\text{ c}\in(0,\pi)$ such that, $\text{f}'(\text{c})=\frac{\text{f}(\pi)-\text{f}(0)}{\pi-0}$

$\Rightarrow\ \cos\text{c}-2\cos2\text{c}=\frac{\sin\pi-\sin2\pi-\sin0+\sin2.0}{\pi-0}$

$\Rightarrow\ 2\cos2\text{c}-\cos\text{c}=\frac{0}{\pi}$

$\Rightarrow\ 2.(2\cos^2\text{c}-1)-\cos\text{c}=0$

$\Rightarrow\ 4\cos^2\text{c}-2-\cos\text{c}=0$

$\Rightarrow\ 4\cos^2\text{c}-\cos\text{c}-2=0$

$\Rightarrow\ \cos\text{c}=\frac{1\pm\sqrt{1+32}}{8}=\frac{1\pm\sqrt{33}}{8}$

$\therefore\ \text{c}=\cos^{-1}\Big(\frac{1\pm\sqrt{33}}{8}\Big)$

Also, $\cos^{-1}\Big(\frac{1\pm\sqrt{33}}{8}\Big)\in(0,\pi)$

Hence, mean value theorem has been verified.

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