# Thread: Using derivative and curve to solve an equation

1. ## Using derivative and curve to solve an equation

I need to solve for all values of a the number of solutions to the equation

$e^{x}=ax$

$e^x=ax \Leftrightarrow \frac {e^x}{x}=a$
We set f(x) = e^x / x and derive, ending up with $\frac {xe^x-e^x}{x^2} = \frac {e^x(x-1)}{x^2}$
meaning that the curve changes at x = 1.

We put up a value? table
x | 0 1
f'(x) | ~ - 0 +
f(x) | ~ e

We look at the limits
$\lim_{x\to 0^+}f(x)=\infty$
$\lim_{x\to +\infty}f(x)=\infty$

So we have a curve with its minimum point at (1,e).
The solutions are:
a < e has 0 solutions
a = e has 1 solution
a > e has 2 solutions

Correct?

Edit:
I forgot to add that we need to check for asymptotes. Since f(x) / x approaches infinity as x->infinity there aren't any asymptotes.

2. Originally Posted by Sabo
I need to solve for all values of a the number of solutions to the equation

$e^{2}=ax$
is this really the equation, or am I missing something?

$y = e^2$ is a constant function

$y = ax$ is a linear function for all $a \in \mathbb{R}$

the two lines intersect only once for all values of $a \ne 0$

if $a = 0$, the two lines are parallel ... no solution.

3. Is this supposed to be
$e^{2x}=ax$

$\displaystyle e^{2x}=ax \Leftrightarrow \frac {e^{2x}}{x}=a$

If I am right, then there's a problem with your derivaitive:
We set f(x) = e^2 / x and derive, ending up with $\displaystyle \frac {x \cdot 2 e^{2x} - e^{2x}}{x^2}$
etc.

-Dan

4. Woa!

I'm sorry. The equation is e^x = ax!

5. Originally Posted by Sabo
The equation is e^x = ax!
The solutions are:
a < e has 0 solutions
a = e has 1 solution
a > e has 2 solutions
for $0 \le a < e$ , no solutions

for $a < 0$ ... one solution

the rest is ok

6. Yeah, seems like I missed the second part of the curve. Thanks.