Show equation has only one solution. Show that the identity $e^{2x/3}(2+x^2)=1$ has only one solution in $\mathbb{R}$.
I thought we can show function is increasing, continuous and has positive and negative values. It’s easy to show that function is increasing and has positive value. Although, it seems obvious function is continuous I cannot prove it formally. Also I am interested how can I find negative value of the function. Can you help me?
 A: Let $f(x)$ be $e^{2x/3}(2+x^2).$ Since composing, adding, or multiplying continuous functions results in a continuous function, $f$ is continuous. The derivative of $f(x)$ is $\frac{(2x^2+6x+4)e^{2x/3}}{3}.$ This derivative is zero when $x = -2$ or $x=-1.$ In both of those cases, $f(x)$ is greater than $1.$ Also, $\lim_{x \to \infty} f(x) = \infty$ and $f(-5)$ is less than $1.$ By the Intermediate Value Theorem, a solution exists.
We prove the rest by contradiction. Assume that the equation has two solutions. Since the derivative of $f(x)$ has finitely many zeros, there are two solutions with no solutions between them. Therefore, there are two solutions $a$ and $b$ with $b>a$ and with no solutions between them such that $a$ is the greatest solution for which two solutions of this form exist.
Assume $f(x) < 1$ between two solutions. By Rolle's theorem, the derivative is $0$ between those two solutions, but the derivative is nonzero if $f(x) \leq 1.$ This is a contradiction. Therefore, between any two solutions, there must be a point where $f(x) \geq 1.$
Case $1$: $f(x) < 1$ between $a$ and $b.$  This is a contradiction.
Case $2$: $f(x) > 1$ between $a$ and $b.$  The derivative must be nonnegative at $a$ and nonpositive at $b$ since the function is greater than $1$ between $a$ and $b.$ The derivative at $b$ cannot be zero. If it is negative, since $\lim_{x \to \infty} f(x) = \infty$, by the Intermediate Value Theorem, there must be another solution greater than $b.$ Choose the smallest such solution and call it $d.$ There are no solutions between $b$ and $c.$ This is a contradiction of the definition of $a$ and $b.$ Therefore, the equation has only one solution.
