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

1 Answer
Dec 3, 2016

Answer:

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]}