Question

i have provided help for you to answer this problem. please input your answers in the boxes as you do for a regular problem. use the graph to help answer the questions. let $f(x)=(x + 1)^2$. a function can only have an inverse if it is one - to - one. each half of the parabola is one - to - one. they are asking for the half that is increasing. give the largest domain on which $f$ is one - to - one and increasing. give the range of $f$. to find the inverse remember you switch the x and the y (f(x)) and then solve the equation for y. find the inverse of $f$ restricted to the domain above. $f^{-1}(x)=$. to answer the following two questions, think about what happens to the x - values of the function compared to the x - values of the inverse function. the above answers for domain and range can help with these questions. give the domain of $f^{-1}$. give the range of $f^{-1}$.

Explanation:

Step1: Find the vertex of the parabola

The function $f(x)=(x + 1)^2$ is in vertex - form $y=a(x - h)^2+k$, where the vertex is $(-1,0)$. A parabola $y=(x + 1)^2$ opens upwards ($a = 1>0$). The function is one - to - one and increasing for $x\geq - 1$.

Step2: Determine the range of $f(x)$

Since the vertex of the parabola $y=(x + 1)^2$ is $(-1,0)$ and it opens upwards, the range of $f(x)$ is $y\geq0$.

Step3: Find the inverse of $f(x)$

Let $y=(x + 1)^2$. Switch $x$ and $y$: $x=(y + 1)^2$. Solve for $y$: $y+1=\sqrt{x}$ (we take the positive square root because we are considering the increasing part of the original function), so $y = f^{-1}(x)=\sqrt{x}-1$.

Step4: Determine the domain and range of $f^{-1}(x)$

The domain of $f^{-1}(x)$ is the range of $f(x)$, so the domain of $f^{-1}(x)$ is $x\geq0$. The range of $f^{-1}(x)$ is the domain of $f(x)$ for the one - to - one and increasing part, so the range of $f^{-1}(x)$ is $y\geq - 1$.

Answer:

1. Domain of $f$: $x\geq - 1$
2. Range of $f$: $y\geq0$
3. Inverse function $f^{-1}(x)$: $\sqrt{x}-1$
4. Domain of $f^{-1}$: $x\geq0$
5. Range of $f^{-1}$: $y\geq - 1$