Question

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

Answer 1

`x = 3" "" "` or `" "" "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 `(x-3)`, `(x+2)`, and `(e^x-x)` 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(x) = 1/((x-3)(x+2)(e^x-x))` will have asymptotes at `x=3` and `x=-2`.

Final Answer