Question

Which function must have a restricted domain in order to have an inverse function?

Answer 1

Answer:

B

Step-by-step explanation:

Consider the function $f(x)=8x^2-6x+10$

Rewrite it as

$f(x)\\ \\=8\left(x^2-\dfrac{3}{4}x+\dfrac{5}{4}\right)\\ \\=8\left(x^2-2\cdot \dfrac{3}{8}\cdot x+\dfrac{5}{4}\right)\\ \\=8\left(x^2-2\cdot \dfrac{3}{8}\cdot x+\dfrac{9}{64}-\dfrac{9}{64}+\dfrac{5}{4}\right)\\ \\=8\left(x-\dfrac{3}{8}\right)^2+142$

The domain of this function is $x\in (-\infty, \infty)$ and the range is $y\in [142,\infty)$

Find the inverse function:

$y-142=8\left(x-\dfrac{3}{8}\right)^2\\ \\x=\pm \sqrt{\dfrac{1}{8}(y-142)}+\dfrac{3}{8}\\ \\f^{-1}(x)= \sqrt{\dfrac{1}{8}(x-142)}+\dfrac{3}{8}$

So, the domain of the inverse function is $x\in [142,\infty)$ and the range will be $\left[\dfrac{3}{8},\infty\right)$

So, the domain must be restricted