Question

Construct graph of the following function:

$y=\{\begin{array}{cc}1-\sqrt{1-x^2},& \text{if}x\le 1\\ 1+{\log }_{\frac{1}{2}}x,& \text{if}x>1\end{array}$

Answer 1

for $y=1-\sqrt{1-x^2}$

since domain for tis region is

$xϵ[-1,1]$ to make $1-x^2\ge 0$

$\frac{dy}{dx}=\frac{2x}{2\sqrt{1-x^2}}\frac{dy}{dx}=0⟹x=0$

$\frac{d^{2}y}{d^{2}x}=\frac{x{(1-x^2)}^{-3/2}}{2\sqrt{1-x^2}}\times (-2x)+\frac{1}{\sqrt{1-x^2}}$

$\frac{d^{2}y}{d^{2}x}=1>0$

so minima is attained

and $y_{x=0}=0$

and $y(-1)=y\left(1\right)=1$

now for part two

$x>1$

$y={\log }_{1/2}x=1-{\log }_{2}x$

for $x\to 1$

$li{m}_{x\to 1}y=1$

and at

$li{m}_{x\to \infty }y=-\infty$

and $y=0=1-{\log }_{2}x⟹{\log }_{2}x=1⟹x=2$

Hence the resultant graph is show above.