Questions: Find the entire domain on which the function f is one-to-one and non-decreasing. Write the domain in interval notation. f(x) = (x+5)^2 [-5, infinity) Find the inverse of f restricted to that domain. f^(-1)(x) = [0, infinity) To help Identify a domain on which the function is non-decreasing, determine an interval on the x-axis for which the function values increase as the x-values increase. Graphing the function may be useful. How does restricting the domain to this interval make the function one-to-one? How does the restricted domain determine which root, positive or negative, should be considered when finding the inverse of the quadratic function?

Find the entire domain on which the function f is one-to-one and non-decreasing. Write the domain in interval notation.
f(x) = (x+5)^2
[-5, infinity)

Find the inverse of f restricted to that domain.
f^(-1)(x) = [0, infinity)

To help Identify a domain on which the function is non-decreasing, determine an interval on the x-axis for which the function values increase as the x-values increase. Graphing the function may be useful. How does restricting the domain to this interval make the function one-to-one? How does the restricted domain determine which root, positive or negative, should be considered when finding the inverse of the quadratic function?

Transcript text: PREVIOUS ANSWERS PRACTICE ANOTHER Find the entire domain on which the function $f$ is one-to-one and non-decreasing. Write the domain in interval notation. \[ \begin{array}{l} f(x)=(x+5)^{2} \\ {[-5, \infty)} \end{array} \] Find the inverse of $f$ restricted to that domain. \[ f^{-1}(x)=[0, \infty) \] $+$ To help Identify a domain on which the function is non-decreasing, determine an interval on the $x$-axis for which the function values increase as the $x$-values increase. Graphing the function may be useful. How does restricting the domain to this interval make the function one-to-one? How does the restricted domain determine which root, positive or negative, should be considered when finding the inverse of the quadratic function?

Solution

Solution Steps

To find the inverse of the function $ f(x) = (x+5)^2 $ restricted to the domain $[-5, \infty)$, we need to ensure that the function is one-to-one and non-decreasing. The function is non-decreasing on this domain because it is a quadratic function that opens upwards. To find the inverse, we solve for $ x $ in terms of $ y $ from the equation $ y = (x+5)^2 $, and then restrict the solution to the non-negative root because the domain $[-5, \infty)$ corresponds to the range $[0, \infty)$.

To solve the problem, we need to find the domain on which the function $ f(x) = (x+5)^2 $ is one-to-one and non-decreasing, and then find the inverse of $ f $ restricted to that domain.

Step 1: Determine the Domain for One-to-One and Non-Decreasing Function

The function $ f(x) = (x+5)^2 $ is a quadratic function, which is a parabola opening upwards. A quadratic function is non-decreasing on the interval from its vertex to infinity. The vertex of the parabola $ f(x) = (x+5)^2 $ is at $ x = -5 $.

Thus, the function is non-decreasing on the interval $[-5, \infty)$.

Step 2: Restrict the Domain

To make the function one-to-one, we restrict the domain to $[-5, \infty)$. On this interval, the function is non-decreasing and passes the horizontal line test, meaning it is one-to-one.

Step 3: Find the Inverse Function

To find the inverse of $ f(x) = (x+5)^2 $ on the restricted domain $[-5, \infty)$, we solve for $ x $ in terms of $ y $:

Start with the equation: \[ y = (x+5)^2 \]
Solve for $ x $: \[ \sqrt{y} = x + 5 \] \[ x = \sqrt{y} - 5 \]

Since the domain of the original function is $[-5, \infty)$, the range of the function is $[0, \infty)$. Therefore, the inverse function is defined for $ y \in [0, \infty) $.

Thus, the inverse function is: \[ f^{-1}(y) = \sqrt{y} - 5 \]