# Number of elements in the set $A=\{x\in\mathbb{R}|f(x)=1\}$.

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?

• It's $+3x^2$, although it doesn't make things simpler. – Arnaud Mortier Jun 16 '18 at 16:39
• Indeed, It was a typo, i edited it – C. Cristi Jun 16 '18 at 16:43

$$3^{x^3-3x}-3^{x+1}+x^3-3x-x=1$$ 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?
• @C.Cristi You mean how to notice it? I noticed that the exponents were kind of similar to the polynomial summands and tried to expose this fact. Then when you arrive at $g(X)=g(Y)$ it is natural to study the variations of the function $g$ to see how it can happen that it takes the same value twice. – Arnaud Mortier Jun 16 '18 at 17:07