# Check if the Mean Value Theorem can be applied to f(x)=4arccos(x) on [1,1]. If so, find all values c in (-1, 1) guaranteed by the Mean Value Theorem. If the Mean Value Theorem does not apply, then justify that conclusion.

Check if the Mean Value Theorem can be applied to $f\left(x\right)=4\mathrm{arccos}\left(x\right)$ on [-1,1]
If so, find all values c in(-1,1) guaranteed by the mean Value theorem. If the Mean Value theorem does not apply, then justify that conclusion
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Carly Yang
$f\left(x\right)=4\mathrm{arccos}x\phantom{\rule{0ex}{0ex}}{f}^{\prime }\left(x\right)=4\left(\frac{-1}{\sqrt{1-{x}^{2}}}\right)\phantom{\rule{0ex}{0ex}}{f}^{\prime }\left(x\right)=\frac{-4}{\sqrt{1-{x}^{2}}}\phantom{\rule{0ex}{0ex}}{f}^{\prime }\left(c\right)=\frac{-4}{\sqrt{1-{c}^{2}}}$
Applying Mean Value Theorem:
${f}^{\prime }\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}\phantom{\rule{0ex}{0ex}}{f}^{\prime }\left(c\right)=\frac{4\mathrm{arccos}1-4\mathrm{arccos}\left(-1\right)}{1-\left(-1\right)}\phantom{\rule{0ex}{0ex}}{f}^{\prime }\left(c\right)=\frac{0-4\left(\pi \right)}{2}\phantom{\rule{0ex}{0ex}}{f}^{\prime }\left(c\right)=-2\pi$
Now
$⇒\frac{-4}{\sqrt{1-{c}^{2}}}=-2\pi \phantom{\rule{0ex}{0ex}}⇒\frac{2}{\sqrt{1-{c}^{2}}}=\pi \phantom{\rule{0ex}{0ex}}⇒\sqrt{1-{c}^{2}}=\frac{2}{\pi }$
square on both sides.
$⇒1-{c}^{2}=\frac{4}{{\pi }^{2}}\phantom{\rule{0ex}{0ex}}⇒{c}^{2}=1-\frac{4}{{\pi }^{2}}\phantom{\rule{0ex}{0ex}}⇒c=±\sqrt{1-\frac{4}{{\pi }^{2}}}$
which gives us c=0.77 and c=-0.77
\So, Mean Value Theorem can be applied.