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

1 Answer
Dec 27, 2016

Answer:

vertical asymptotes at x = 1, x= 1.1 , x = 2

Explanation:

The denominator of f(x) cannot be zero as this would make f(x) undefined. Equating the denominator to zero and solving gives the values that x cannot be and if the numerator is non-zero for these values then they are vertical asymptotes.

solve: #(x-2)(x-1)(e^x-3)=0#

#rArrx=1,x=2,x=ln3" are the asymptotes"#
graph{1/((x-2)(x-1)(e^x-3) [-10, 10, -5, 5]}