Exercises: Restrict Domain for Invertibility

Which of the following functions fails the Horizontal Line Test on its full domain?

When you restrict the domain of $f(x) = x^2$ to $x \geq 0$, what changes and what stays the same?

The function $f(x) = (x - 3)^2$ has its vertex at $x = 3$. Which domain restriction makes $f$ one-to-one?

$f(x) = x^2$ with domain $\mathbb{R}$ has no inverse. If we restrict to $x \geq 0$, the inverse is $f^{-1}(x) = \sqrt{x}$. Why does the restriction fix the problem?

The function $f(x) = (x - 4)^2$ has its vertex at $x = 4$. To restrict the domain so that $f$ is one-to-one on the right branch, enter the smallest value of $x$ allowed (as a number).

$f(x) = x^2$ is restricted to $x \geq 0$. What is $f^{-1}(x)$?

$f(x) = x^2 + 2$ is restricted to $x \geq 0$. The range of this restricted function is $[a, \infty)$. What is $a$?

$f(x) = x^2 + 2$, restricted to $x \geq 0$. Find $f^{-1}(x)$ by setting $y = x^2 + 2$, swapping variables, and solving for $y$ (taking the positive root). Enter your answer as a function rule — type: sqrt(x-2).