Question

Which statement describes the inverse of m(x) = x^2 – 17x?

Answer 1

Given:

The function is

$m(x)=x^2-17x$

To find:

The inverse of the given function.

Solution:

We have,

$m(x)=x^2-17x$

Substitute m(x)=y.

$y=x^2-17x$

Interchange x and y.

$x=y^2-17y$

Add square of half of coefficient of y , i.e., $\left(\dfrac{-17}{2}\right)^2$ on both sides,

$x+\left(\dfrac{-17}{2}\right)^2=y^2-17y+\left(\dfrac{-17}{2}\right)^2$

$x+\left(\dfrac{17}{2}\right)^2=y^2-17y+\left(\dfrac{17}{2}\right)^2$

$x+\left(\dfrac{17}{2}\right)^2=\left(y-\dfrac{17}{2}\right)^2$        $[\because (a-b)^2=a^2-2ab+b^2]$

Taking square root on both sides.

$\sqrt{x+\left(\dfrac{17}{2}\right)^2}=y-\dfrac{17}{2}$

Add $\dfrac{17}{2}$ on both sides.

$\sqrt{x+\left(\dfrac{17}{2}\right)^2}+\dfrac{17}{2}=y$

Substitute $y=m^{-1}(x)$.

$m^{-1}(x)=\sqrt{x+(\dfrac{189}{4}})+\dfrac{17}{2}$

We know that, negative term inside the root is not real number. So,

$x+\left(\dfrac{17}{2}\right)^2\geq 0$

$x\geq -\left(\dfrac{17}{2}\right)^2$

Therefore, the restricted domain is $x\geq -\left(\dfrac{17}{2}\right)^2$ and the inverse function is $m^{-1}(x)=\sqrt{x+(\dfrac{189}{4}})+\dfrac{17}{2}$.

Hence, option D is correct.

Note: In all the options square of $\dfrac{17}{2}$ is missing in restricted domain.