Question

Let f(x)=2x^2-16. Find f^-1(x)

Answer 1

as you already know, to get the inverse of any expression we start off by doing a quick switcheroo on the variables and then solving for "y", let's do so.

`\stackrel{f(x)}{y}~~ = ~~2x^2-16\hspace{5em} \stackrel{\textit{quick switcheroo}}{x~~ = ~~2y^2-16} \\\\\\ x+16=2y^2\implies \cfrac{x+16}{2}=y^2\implies \cfrac{x}{2}+\cfrac{16}{2}=y^2 \\\\\\ \cfrac{x}{2}+8=y^2\implies {\Large \begin{array}{llll} \pm\sqrt{\cfrac{x}{2}+8}~~ = ~~\stackrel{f^(-1)(x)}{y} \end{array}}`

Answer 2

`f^(-1)(x)=\pm \sqrt{(x)/(2)+8}, \quad \;\;\;x\geq -16`

Explanation:

Given function:

`f(x)=2x^2-16`

The given function has an unrestricted domain and a restricted range.

• Domain of f(x): (-∞, ∞)
• Range of f(x): [-16, ∞)

Inverse of a function

f⁻¹(x) is the notation for the inverse of the function. The inverse of a function is its reflection in the line y = x.

To find the inverse of the given function, swap f(x) for y:

`\implies y=2x^2-16`

Rearrange the equation to make x the subject:

`\implies y+16=2x^2-16+16`

`\implies y+16=2x^2`

`\implies (y+16)/(2)=(2x^2)/(2)`

`\implies (y+16)/(2)=x^2`

`\implies x^2=(y)/(2)+8`

`\implies √(x^2)=\sqrt{(y)/(2)+8}`

`\implies x=\pm \sqrt{(y)/(2)+8}`

Swap the x for f⁻¹(x) and the y for x:

`\implies f^(-1)(x)=\pm \sqrt{(x)/(2)+8}}`

The domain of the inverse function is the range of the function.

Therefore, the domain of the inverse function is restricted:

• Domain of f⁻¹(x): [-16, ∞)

Solution

`f^(-1)(x)=\pm \sqrt{(x)/(2)+8}, \quad \;\;\;x\geq -16`