Question

If $f(x)=\left[\begin{array}{ccc}x & , & x<1 \\ x^2 & , & 1 \leq x \leq 4 \\ 8 \sqrt{x} & , & x>4\end{array}\right.$, then find $f^{-1}(x)$.

Answer 1

SOLUTION : $$

$\text { Case I } y=x \quad x<1 \\$

$x=y \quad y<1 \\$

$f^{-1}(x)=x \quad x<1 \\$

$\text { Case II } y=x^2 \quad 1 \leq x \leq 4 \\$

$x^2=y \quad 1 \leq y \leq 16 \\$

$x=\sqrt{y} 1 \leq y \leq 16 \\$

$f^{\prime}(x)=\sqrt{x}  1 \leq x \leq 16$

Case III $\quad y=8 \sqrt{x} \quad x>4$

$x=\frac{y^2}{64} \quad y>16 ; \quad f^{-1}(x)=\frac{x^2}{64} \quad x>16$