Find amount of roots

Question

We are given equation $$\frac {e^x}{x^2} = a$$.
Task is to find how many solutions equation would have depending on values of a.
Let's illustrate a(x):

It's easy to conclude that there aren't no solutions when $a < 0$.
Also it's obvious that a(x) has extremal point at $x = 2; a(2)= \frac {e^2}{4}$.
So:
- $a(x)$ grows at interval $(-\infty, 0)$ from 0 to $+\infty$
- $a(x)$ decreases at interval $(0, 2)$ from $+\infty$ to $\frac {e^2}{4}$
- $a(x)$ grows at interval $(2, +\infty)$ from $\frac {e^2}{4}$ to $+\infty$

It's easy to understand that f(x) would have 2 roots with parameter $a=\frac {e^2}{4}$.
But how would I find amount of roots for $(0, 2)$, $(2, +\infty)$?
Answer says they are 1 and 3, accordingly. But I see no ways of proving it. I would like a pointer. Thank you for time, guys.

Answer 1

Well you only need to find the parametric solutions of the system \begin{cases} y=\frac{e^x}{x^2}\\ y=a \end{cases} The best way is to draw a simple sketch of the curve $y=\frac{e^x}{x^2}$, and next take a look at the possible intersections with lines parallel to $x$-axis.

From the sketch above we can state that: \begin{cases} \forall a \lt 0\to 0\text{ solutions}\\ \forall a | 0 \lt a \lt \frac{e^2}{4} \to 1 \text{ solution}\\ a = \frac{e^2}{4} \to 2 \text{ coincident solutions}\\ \forall a | a \gt \frac{e^2}{4} \to 3 \text{ different solutions}\\ \end{cases} You can easily find point $A$ by letting $f'(x) = D[f(x)] = D[\frac{e^x}{x^2}] = \frac{e^x (-2+x)}{x^3} = 0\Rightarrow x=2$. There are no relative minima (except $x=2$), nor maxima, as $f'(x)=0$ only if $x=2$, so there are no extra stationary points where a line parallel to $x$-axis can intersect $f(x)$. Moreover, notice that $f'(x) \gt 0 \iff x \lt 0 \lor x \gt 2$. Thus $f(x)$ is an increasing function in $]-\infty;0]$ and in $]2;+\infty[$, which means you have ($\forall a \gt \frac{e^2}{4}$) necessarily $1$ solution of the equation $\frac{e^x}{x^2}=a$ in $]-\infty;0]$, the second one in $]0;2[$, and a final one in $]2;+\infty[$, for a total of $3$ different solutions. A similar reasoning can be carried out when $0 \lt a \lt \frac{e^2}{4}$, to find the exact number of solutions as written above.

Answer 2

take $e^x=ax^2$ and do cases based on a, it should be much easier

Answer 3

Analyse $e^x$ and $ax^2$.

Hand made graphs will suffice.

Clearly, no solution for non-positive $a$. For very small positive $a$, the graph of $e^x$ will intersect at a negative point and go directly above to infinity and beyond resulting in one solution.

With significantly large $a$, the graph will intersect it in $3$ points, for some $x$, $e^x$ will be below it. Obviously, one point will come when $2$ solutions are available and $e^x$ touches $ax^2$. After it, $3$ solutions exist and before it $2$.

How to find that $a$ :

$$e^X=aX^2=2aX$$

You can solve it easily. I have used equal value and slope at a point $X$.