Question

If the function $f:[1,\infty )\to [1,\infty )$ is defined by $f\left(x\right)=3^{x(x-1)}$ then $f^{-1}\left(x\right)$ is

• A. ${\left(\frac{1}{2}\right)}^{x(x-1)}$
• B. $\frac{1}{2}(1+\sqrt{1+4{\log }_{2}x})$
• C. $\frac{1}{2}(1-\sqrt{1+4{\log }_{2}x})$
• D. not defined

Answer 1

The correct option is B $\frac{1}{2}(1+\sqrt{1+4{\log }_{2}x})$

For the given domain, the function is one one.

Thus,

$f\left(x\right)=2^{x(x-1)}$

$y=2^{x^2-x}$

$lny=(x^2-x)ln2$

$lo{g}_{2}y=x^2-x$

$lo{g}_{2}y=(x-\frac{1}{2}{)}^2-\frac{1}{4}$

$lo{g}_{2}y+\frac{1}{4}=(x-\frac{1}{2}{)}^2$

$\frac{\pm 1}{2}\left(\sqrt{1+4lo{g}_{2}y}\right)=x-\frac{1}{2}$

Since $f:[1,\infty )\to [1,\infty )$

Thus domain is $[1,\infty )$

Hence domain is positive so negative sign will be ignored. $So$

$x-\frac{1}{2}=\frac{1}{2}\left(\sqrt{1+4lo{g}_{2}y}\right)$

$x=\frac{1}{2}[1+\sqrt{1+4lo{g}_{2}y}]$

Replacing x and y, gives us

$f^{-1}\left(x\right)=\frac{1}{2}[1+\sqrt{1+4lo{g}_{2}x}]$