# For what values of x, if any, does f(x) = 1/((x-3)(x+2)(e^x-x))  have vertical asymptotes?

Oct 8, 2017

$x = 3 \text{ "" }$ or $\text{ "" } x = - 2$

#### Explanation:

The function will have a vertical asymptote wherever the denominator is $0$ and the numerator is not also $0$.

In this case, our job is a little easier since we don't have to check whether or not the numerator is $0$, since it is always $1$.

The denominator will be $0$ whenever any of its factors $\left(x - 3\right)$, $\left(x + 2\right)$, and $\left({e}^{x} - x\right)$ are $0$. So, let's set each of them equal to $0$ and solve for $x$.

$x - 3 = 0$

color(red)(x = 3

$x + 2 = 0$

color(red)(x = -2

${e}^{x} - x = 0$

${e}^{x} = x$

Hmm... this might be trickier. Does ${e}^{x} = x$ have any solutions?

Let's consider if $x$ is negative. Since ${e}^{x}$ is always positive, it cannot equal $x$ if $x$ is negative. So we have no negative solutions.

If $x$ is 0, then ${e}^{x} = 1$, so that is not a solution.

This leaves us with the fact that $x$ could be positive, but we need to consider how fast the functions are increasing.

For any value of $x$, ${e}^{x}$ is increasing at a rate of ${e}^{x}$, and $x$ is increasing at a rate of $1$. For positive values of $x$, ${e}^{x} > 1$. This means that for positive values of $x$, ${e}^{x}$ increases faster than $x$.

Since ${e}^{x}$ starts out (at $x = 0$) greater than $x$, and continues to grow faster than it, it will ALWAYS be greater than $x$ for any value, since the function $y = x$ can never "catch up" to it.

Therefore, there are no positive solutions to ${e}^{x} = x$.

This means there are no solutions to ${e}^{x} = x$.

Therefore, the function $f \left(x\right) = \frac{1}{\left(x - 3\right) \left(x + 2\right) \left({e}^{x} - x\right)}$ will have asymptotes at $x = 3$ and $x = - 2$.