The minimum value of $\mathrm{k \in \Bbb Z}$ for the equation $\mathrm{e}^x = kx^2$ to have maximum number of solutions

Question

Find the minimum value of $k\in\Bbb Z$ for the equation $\mathrm e^x = kx^2$ to have the maximum number of solutions?

I don't get how to even get going. Any hint would do.

Answer 1

It is easier to solve the question this way:

$e^x = kx^2 \implies \dfrac{e^x}{x^2}= k$

We want this equation to have $3$ solutions.

For $f(x)= \dfrac {e^x}{x^2}$

$f'(x)= \dfrac{(x-2)e^x}{x^3}$

Clearly, $f(x)$ is increasing for $x>2$ , decreasing for x $\in (0,2)$ and increasing for $x<0$

$\lim _{x\to \infty} f(x) = \infty$

$\lim _{ x \to - \infty}f(x)= 0$

Also, $x= 0$ is the vertical asymptote.

This information is sufficient to plot a rough sketch of $f(x)$

$x=2$ is a point of local minima (no need to check $f''(x)$, just notice that f(x) is decreasing then increasing around 2 so it has to be a point of minima ).

Now note that $f(2) = e^2/ 4 \approx 1.8$

We can clearly see from the graph that any horizontal line intersecting the graph above y-value $1.8$ will intersect it maximum number of times = $3$ times.

Hence the minimum value of $k \in \mathbb Z = 2$

Answer 2

It is clear there are no solutions for $k\le 0$. For $k\ge 1$, there is always a solution with $x<0$

Now let us show that there can be at most $3$ solutions in total, no matter what $k$ is. By Rolle's theorem, between any two zeroes of $e^{x}-kx^2$ there lies a zero of the derivative $e^{x}-2kx$. (and the zeroes are simple for integer $k$). Between any two zeroes of those lies a zero of the second derivative $e^x-2k$. That has exactly one zero if $k>0$, so there can be at most $3$ zeroes.

Now we examine $k=1$ and $k=2$. For $k=1$, $$e^x-x^2\ge 1+x-\frac{x^2}{2}+\frac{x^3}6.$$ That is positive at $0$. The derivative is $$1-x+\frac{x^2}2, $$ which is always positive so there are no solutions of $e^x=x^2$ in positive $x$.

Now all we need to show is that for $k=2$ there are two positive solutions, so three in total. It is enough to note that $e^2-2\times2^2<0$ and that $e^0>0$ and $\lim_{x\to\infty} e^x-2x^2>0$, then we get two positive solutions by the intermediate value theorem.