What is the difference between $f(x)=x^2 +1$ and $f(x)=x^3 -1$ when finding the inverse?

Question

I'm doing some exercises on computing the inverse of each function.

In exercise number 56 I did an example where I have to compute the inverse of the function. With my understanding $f(x)=x^2 +1$ inverse is $f^{-1}(x)=\sqrt{x-1}$. The Solutions Book says $f^{-1}(x)$ does not exist.

On exercise 60 I got almost the same concept to get the inverse. $f(x)=x^3-1$ which my answer is $f^{-1}(x)=\sqrt[3]{x+1}$. The solutions Book says $f^{-1}(x)=\sqrt[3]{x+1}$ which I was right.

My question is why is exercise 56 inverse of function does not exist and on exercise 60 it does?

Answer 1

Think what happens e.g. at the point $x=-3$ for the 1st function and the 1st inverse function. That's what the book means. The 1st function is not reversible when viewed over the whole real line.

Answer 2

The solutions of $y=x^2+1$ are $x=\sqrt{y-1}$ and $x=-\sqrt{y-1}$. There are two solutions, this function is not bijective, there is no unique solution $x$ given $f(x)$.

Answer 3

Here is a graph of $f(x) = x^2 + 1$ (green) and the inverted graph (blue, light blue). The inverted graph is a relation but no function, because it assigns more than one value for each argument $x$ with $x > 1$. So there is no inverse function.

And here is the graph of $f(x) = x^3 - 1$ (green) and the inverted graph (blue). In this case the inverted graph is a function and the inverse function exists.

Answer 4

To solve for the inverse of $f(x) = x^3 - 1$, we set $y = x^3 - 1$ and solve for $x$ in terms of $y$.
\begin{align*} y & = x^3 - 1\\ y + 1 & = x^3\\ \sqrt[3]{y + 1} & = x \end{align*} Since each real number has a unique cube root, $x$ is uniquely determined by the value of $y$. Hence, $f^{-1}(x) = \sqrt[3]{x + 1}$ exists.

If we attempt to do the same thing with the function $f(x) = x^2 + 1$, we obtain \begin{align*} y & = x^2 + 1\\ y - 1 & = x^2\\ \sqrt{y - 1} & = |x| && \text{since $\sqrt{x^2} = |x|$}\\ \pm \sqrt{y - 1} & = x \end{align*} For each $y > 1$, there are two values of $x$ such that $y = x^2 + 1$. Hence, $x$ is not uniquely determined by $y$. Thus, the equation $x = \pm \sqrt{y - 1}$ does not define a function, so the inverse of $f(x) = x^2 + 1$ does not exist.

Remember that if $f$ is a function of $x$, there is a unique $y$-value for each $x$-value in the domain. This means that a vertical line will cross the graph of a function at most once, which is known as the Vertical Line Test.

A function has an inverse if in addition to being able to express $y$ as a function of $x$, we can write $x$ as a function of $y$. For this to occur, not only must there be a unique value of $y$ for each value of $x$, there must be a unique value of $x$ for each value of $y$. This means that a horizontal line will cross the graph of a function at most once, which is known as the Horizontal Line Test.

Observe that for each value of $y > 1$, a horizontal line crosses the graph of $f(x) = x^2 + 1$ twice, so $x$ is not uniquely determined by $y$ if $y > 1$. On the other hand, for each value of $y$, a horizontal line crosses the graph of $f(x) = x^3 - 1$ once, so $x$ is uniquely determined by $y$ for each value of $x$.

Observe that for each $x > 1$, a vertical line crosses the graph of $y = \pm \sqrt{x^2 - 1}$ twice, so the equation $y = \pm\sqrt{x^2 - 1}$ does not represent the graph of a function. On the other hand, for any value of $x$, a vertical line crosses the graph of $y = x^3 - 1$ exactly once, so its graph does represent a function.