Find f.
f ' (x)= 4/{square root(1-2[squared])}, f(1/2)= 1
Any insight as to how to handle this would be wonderful
Hello, CALsunshine!
I assume there is a typo . . .
$\displaystyle f'(x)\:=\:\frac{4}{\sqrt{1-x^2}},\quad f\left(\frac{1}{2}\right)\:=\: 1$
Find $\displaystyle f(x).$
Integrate: . $\displaystyle f(x) \;=\;\int\frac{4}{\sqrt{1-x^2}}\,dx \quad\Rightarrow\quad f(x) \;=\;4\sin^{\text{-}1}(x) + C$
Since $\displaystyle f\left(\frac{1}{2}\right) \,=\,1$, we have: .$\displaystyle 1 \;=\;4\sin^{\text{-}1}\left(\frac{1}{2}\right) + C$
. . Then: .$\displaystyle 1 \;=\;4\left(\frac{\pi}{6}\right) + C \quad\Rightarrow\quad C \:=\:1 - \frac{2\pi}{3}$
Therefore: .$\displaystyle f(x)\;=\;4\sin^{\text{-}1}(x) + 1 - \frac{2\pi}{3}$