Equation solving from a given function relationship

Question

I have a function $f:\mathbb{R}\rightarrow\mathbb{R}$ which can derives. $f$ is strictly decreasing at $(-\infty,-1]$ and strictly increasing at $[-1,+\infty)$ with $f(-1)=-16$. Now I want to solve the equation: $$f(2e^x-x)=f(x^2-x+2)$$ I can notice that $x=0$ is an obvious solution but how can I prove that is the only one?

Answer 1

You have $2e^x-x>0\;\forall\,x\in\mathbb{R}$, and $x^2-x+2=(x-1/2)^2+7/4>0\;\forall\,x\in\mathbb{R}.$ It follows that your two expressions are always going to be in the strictly increasing portion of $f$'s domain. Thus, $f$ is one-to-one for the expressions you're equating. Therefore, solving the equation amounts to solving \begin{align*} 2e^x-x&=x^2-x+2\quad\text{or} \\ 2e^x&=x^2+2. \end{align*} Let $g(x)=2e^x-x^2-2$, so that solutions of the above equation are tantamount to finding roots of $g(x)$. Differentiating $g(x)$ once yields $$g'(x)=2e^x-2x,\quad\text{or}\quad g'(x)=2(e^x-x). $$ As mentioned before, this is always strictly positive for all $x$, which implies $g(x)$ is strictly monotonically increasing, and hence can have only one root. As you have already found this root at the origin, there can be no other roots.