TITLE: Number of elements in the set $A=\{x\in\mathbb{R}|f(x)=1\}$.
QUESTION [6 upvotes]: Let $f:\mathbb{R}\to\mathbb{R},f(x)=3^{x^3-3x}-3^{x+1}+x^3-4x$ and $A=\{x\in\mathbb{R}|f(x)=1\}$ if $|A|=$number of elements in A, then $|A|=?$.

I tried differentation and got $$f'(x)=(3x^2-3)3^{x^3-3x}\ln(3)-3^{x+1}\ln(3)+3x^2-4$$
However trying to find a maxima or minima from this expression is really hard... so then there must be some clever way too look at $f$ maybe something with the composition of $2$ functions? any hints?

REPLY [10 votes]: $$3^{x^3-3x}-3^{x+1}+x^3-3x-x=1$$
can be re-written as
$$3^{X}+X=3^{Y}+Y$$
where $X=x^3-3x$ and $Y=x+1$.
Now the function $x\mapsto 3^x+x$ is strictly increasing, therefore injective, so we must have $X=Y$.
So the answer is the number of solutions to $x^3-3x=x+1$. Can you take it from here?