$f (x)=\exp(x/2)−25x^2$. Show that f on $ (4\log(20), \infty)$ has exactly one root.

Question

Let $$f (x)=\exp(x/2)−25x^2$$ Show that $f$ on $ (4\log(20), \infty)$ has exactly one root $x^*$. (Note that log the natural logarithm)

I'm struggling with this question, we were given a hint, which was:

Start by noting that

$$f(x)=\bigg( \exp \bigg( \frac{x}{4} \bigg)−5x \bigg)\bigg(\exp\bigg(\frac{x}{4}\bigg)+5x \bigg)$$

However, I'm not sure what to make of this hint. I started with the fact that $h(x) = g(x)-x$ for a fixed point but not sure where to go from there.

Answer 1

For positive $x$, the function $exp(\frac{x}{4})+5x$ is obviously positive, so the only possibility is that $g(x):=exp(\frac{x}{4})-5x$ has a root.

The derivate of $exp(\frac{x}{4})-5x$ is $\frac{1}{4}exp(\frac{x}{4})-5$, which is positive for $x>4log(20)$ because the inequality $\frac{1}{4}exp(\frac{x}{x})-5>0$ can be transformed into the inequality $x>4log(20)$. So, there is at most one root for $x>4log(20)$. Because of $g(17)<0\ ,\ g(18)>0$, the only root is between $17$ and $18$. $$17.99902115409395014389893303$$ is the numerical value of the root.