Question

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

a. The domain restriction x ≥ StartFraction 17 Over 2 EndFraction results in m–1(x) =StartFraction 17 Over 2 EndFraction minus StartRoot x + StartFraction 289 Over 4 EndFraction EndRoot .


b. The domain restriction x ≥ StartFraction 17 Over 2 EndFraction results in m–1(x) =StartFraction 17 Over 2 EndFraction + StartRoot x + StartFraction 289 Over 4 EndFraction EndRoot .


c. The domain restriction x ≥ Negative StartFraction 17 Over 2 EndFraction results in m–1(x) =StartFraction 17 Over 2 EndFraction minus StartRoot x + StartFraction 289 Over 4 EndFraction EndRoot .


d.The domain restriction x ≥ Negative StartFraction 17 Over 2 EndFraction results in m–1(x) =StartFraction 17 Over 2 EndFraction + StartRoot x + StartFraction 289 Over 4 EndFraction EndRoot .

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((-17)/(2)\right)^2` on both sides,

`x+\left((-17)/(2)\right)^2=y^2-17y+\left((-17)/(2)\right)^2`

`x+\left((17)/(2)\right)^2=y^2-17y+\left((17)/(2)\right)^2`

`x+\left((17)/(2)\right)^2=\left(y-(17)/(2)\right)^2`
`[\because (a-b)^2=a^2-2ab+b^2]`

Taking square root on both sides.

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

Add
`(17)/(2)` on both sides.

`\sqrt{x+\left((17)/(2)\right)^2}+(17)/(2)=y`

Substitute
`y=m^(-1)(x)`.

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

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

`x+\left((17)/(2)\right)^2\geq 0`

`x\geq -\left((17)/(2)\right)^2`

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

Hence, option D is correct.

Note: In all the options square of
`(17)/(2)` is missing in restricted domain.