Question

If $f:(-\infty ,3]\to [7,\infty );f\left(x\right)=x^2-6x+16$, then which of the following is true?

• A. $f^{-1}\left(x\right)=3+\sqrt{x-7}$
• B. $f^{-1}\left(x\right)=3-\sqrt{x-7}$
• C. $f^{-1}\left(x\right)=\frac{1}{x^2-6x+16}$
• D. $f$ is many-one

Answer 1

Answer: (A)

$f^{-1}\left(x\right)=3+\sqrt{x-7}$

Let $f\left(x\right)=y$

$=x^2-6x+16$

$=(x-3{)}^2+7$

So $(x-3{)}^2=y-7$

$\Rightarrow x=3+\sqrt{y-7}$

now $f\left(x\right)=y$

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

So $f^{-1}\left(y\right)=3+\sqrt{y-7}$

Since x,y are arbitrary variables replace y with x to get Option A as the answer.