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I am asked to find the inverse for

$f(x)=x^3+\arctan(x+1)$

I graphed it on Desmos and found that it is one-to-one, so I assumed it does not have an inverse. Is my approach correct?

Raffaele
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Hatem Chalak
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  • Your title conflicts with the body. The body says it is one to one and therefore does not have an inverse. – Ross Millikan Oct 14 '17 at 20:54
  • Because I thought it wouldn't have an inverse if it was one-to-one, but I asked to clarify. Sorry for the confusion that I might have created. The person who answered my question below clarified it for me – Hatem Chalak Oct 14 '17 at 20:58

2 Answers2

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To be sure compute the derivative

$$f'(x)=3x^2+\frac{1}{1+(1+x)^2}$$ which is the sum of two positive quantities so it's positive on all domain $\mathbb{R}$

So the function is injective (technical for $1-1$)

to be invertible it must be also surjective which means that the range is all the co-domain. In other words all $y\in\mathbb{R}$ have an $x$ such that $y=f(x)$.

There are no "holes" in the co-domain $y$-axis

And it happens for this function thanks to the $x^3$ part, because the arctan part gives results only in $(-\pi/2;\;\pi/2)$

The graph of the inverse is the symmetric of the function wrt the line $y=x$

Hope this helps $$ . $$

enter image description here

Raffaele
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I have seen two definitions of one-to-one for a function. Sometimes it means that the function is an injection, that every element in the range is the image of at most one element of the domain. That means if you have $f(x)=f(y)$ then $x=y$. This is the most common version and is reflected in the Wikipedia article. I have also heard it used to mean bijection, which means both an injection and a surjection. I think that version passed out of usage long ago. For a function to have an inverse, it needs to be a bijection, so it needs to be one-to-one under either definition.

Ross Millikan
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