Question

For what values of x, if any, does `f(x) = 1/((x-3)(x+4)(e^x-x))` have vertical asymptotes?

Answer 1

The graph depicts the answer: x = 3 and `x=-4`.

Explanation:

`e^x-x` has the minimum 1, at x = 0.

So, `y to +-oo`, as `x to 3 and x to -4`.

And so, the vertical asymptotes are given by x = 3 and `x = - 4`.

The y-intercept of the graph is `-1/12` and it is the local minimum.

graph{y(x-3)(x+4)(e^x-x)-1=0 [-5, 5, -2.5, 2.5]}