Question

Let $f:\left(-\infty ,\frac{1}{2}\right]\to \left[\frac{1}{2},\infty \right)$ be a function defined as $f(x)=16^{x^2-x}$ then which one of the following must be CORRECT?

• A. $f^{-1}(x)$ is not possible because $f(x)$ is not a bijective function
• B. $f^{-1}(x)=\frac{1}{2}-\sqrt{{\log }_{16}x+\frac{1}{4}}$
• C. $f^{-1}(x)=\frac{1}{2}+\sqrt{{\log }_{16}x+\frac{1}{4}}$
• D. $f^{-1}(x)=\frac{1}{2}+\sqrt{{\log }_{16}x-\frac{1}{4}}$

Answer 1

Answer: (B)

$f(x)=16^{x^2-x}=16^{{\left(x-\frac{1}{2}\right)}^2-\frac{1}{4}}\ge \frac{1}{2}\Rightarrow f(x)$ is onto
$f^{\prime }(x)=(2x-1)16^{x^2-x}\ln 16<0\forall x<\frac{1}{2}$
$\Rightarrow f(x)$ is one-one
$y=16^{x^2-x}\Rightarrow {\left(x-\frac{1}{2}\right)}^2-\frac{1}{4}={\log }_{16}y$
$x=\frac{1}{2}-\sqrt{{\log }_{16}y+\frac{1}{4}}$ OR $x=\frac{1}{2}+\sqrt{{\log }_{16}y+\frac{1}{4}}$ (rejective)
$f^{-1}(x)=\frac{1}{2}-\sqrt{{\log }_{16}x+\frac{1}{4}}.]$