Question

graph the following functions. create a limit on the domain for f(x) so that you can find and graph the inverse function

1. f(x)=x² + 5 f⁻¹(x)=√(x - 5)

domain of f(x)=0,∞) domain of f⁻¹(x)=5,∞)
range of f(x)=5,∞) range of f⁻¹(x)=0,∞)
what limit do we need to put on the domain so that the inverse is a function?
x≥0

Explanation:

Step1: Recall function - inverse property

A function \(y = f(x)\) has an inverse that is also a function if and only if the original function is one - to - one. For the quadratic function \(f(x)=x^{2}+5\), it is a parabola opening upwards with vertex at \((0,5)\). To make it one - to - one, we can restrict the domain.

Step2: Determine domain restriction

If we consider \(x\geq0\), the function \(y = x^{2}+5\) is increasing. Let \(y=x^{2}+5\), then to find the inverse, we solve for \(x\):
\[y=x^{2}+5\Rightarrow x^{2}=y - 5\Rightarrow x=\sqrt{y - 5}\] (since \(x\geq0\)), so \(f^{-1}(x)=\sqrt{x - 5}\). The domain of \(f(x)\) when restricted to \(x\geq0\) gives a range of \(y\geq5\). For the inverse function \(f^{-1}(x)=\sqrt{x - 5}\), the domain is \(x\geq5\) and the range is \(y\geq0\).

Answer:

The domain of \(f(x)\) should be restricted to \(x\geq0\) so that the inverse \(f^{-1}(x)=\sqrt{x - 5}\) is a function.