# find range of f(x)=3sin(root(pi2/16) -x2)

the given function is $f\left(x\right)=3.\mathrm{sin}\left(\sqrt{\frac{{\pi }^{2}}{16}-{x}^{2}}\right)$
since the quantity within the square root can not be negative.
therefore
$\frac{{\pi }^{2}}{16}-{x}^{2}\ge 0\phantom{\rule{0ex}{0ex}}{x}^{2}\le \frac{{\pi }^{2}}{16}\phantom{\rule{0ex}{0ex}}-\frac{\pi }{4}\le x\le \frac{\pi }{4}$
the minimum value of the function is $3.\mathrm{sin}0=0$
and the maximum value of the function is $3\mathrm{sin}\frac{\pi }{4}=\frac{3}{\sqrt{2}}$

therefore range is $\left[0,\frac{3}{\sqrt{2}}\right]$
hope this helps you

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