Question

10.) the function is one-to-one. find its inverse
$f(x)=sqrt{x + 5}$

a) $f^{-1}(x)=x^2 - 5, xgeq0$
b) $f^{-1}(x)=sqrt{x - 5}$
c) $f^{-1}(x)=x^2 + 5, xgeq0$
d) $f^{-1}(x)=(x + 5)^2$

Explanation:

Step1: Replace \( f(x) \) with \( y \)

We start with the function \( f(x)=\sqrt{x + 5} \). Replace \( f(x) \) with \( y \), so we have \( y=\sqrt{x + 5} \).

Step2: Swap \( x \) and \( y \)

To find the inverse, we swap the roles of \( x \) and \( y \). This gives us \( x=\sqrt{y + 5} \).

Step3: Solve for \( y \)

First, square both sides of the equation to eliminate the square root. Squaring both sides, we get \( x^{2}=y + 5 \) (since \( (\sqrt{a})^{2}=a \) for \( a\geq0 \)). Then, subtract 5 from both sides to solve for \( y \). So, \( y=x^{2}-5 \).

Step4: Determine the domain of the inverse function

The original function \( f(x)=\sqrt{x + 5} \) has a range of \( y\geq0 \) (because the square root of a non - negative number is non - negative). The domain of the inverse function is the range of the original function. So, the domain of \( f^{-1}(x) \) is \( x\geq0 \). So the inverse function is \( f^{-1}(x)=x^{2}-5,x\geq0 \).

Answer:

a) \( f^{-1}(x)=x^{2}-5,x\geq0 \)