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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):

graph 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.

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It's better to analyse e^x and ax^2 and see points of intersection. –  Awesome May 7 at 12:58
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3 Answers 3

up vote 3 down vote accepted

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. enter image description here

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.

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

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@Awesome I see your point. It really was very easy to prove with graphs, generated by wolframalpha. It feels like cheating, though. Actually, it is. Then, approaching it analytically, it became really not obvious, that in the one case it's 2 roots and 3 in the another: $4e^{x-2} = x^2$ $2e^{x-2} = x^2$. –  wf34 May 7 at 13:11
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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 $X$.

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