How do you find the vertical, horizontal or slant asymptotes for #f(x)= (3e^(x))/(2-2e^(x))#?

Question

Anna Allen · Accepted Answer

Vertical asymptote is #x=0# and

two horizontal asymptotes #x=0# and #x=-3/2# Vertical asymptote is given by putting denominator equal to zero. Hence it is given by #2-2e^x=0# or #2e^x=2# or #e^x=1# or #x=ln1=0# For horizontal asymptote, let us find out #Lt_(x->oo)(3e^x)/(2-2e^x)# Dividing numerator by #e^x# this is equal to #Lt_(x->oo)3/(2e^-x-2)=-3/2# as #e^-oo=0# Also #Lt_(x->-oo)(3e^x)/(2-2e^x)=0# Hence we have two horizontal asymptotes #x=0# and #x=-3/2# graph{3e^x/(2-2e^x) [-10, 10, -5, 5]}

Caleb Alexander · Answer

undefined To find the vertical asymptotes of the function $ f(x) = \frac{3e^x}{2 - 2e^x} $, we need to identify the values of $ x $ for which the denominator equals zero. Setting the denominator equal to zero, we solve for $ x $:

$$ 2 - 2e^x = 0 $$

$$ 2 = 2e^x $$

$$ 1 = e^x $$

$$ x = \ln(1) $$

Since $ \ln(1) = 0 $, there is no vertical asymptote for this function.

To find the horizontal asymptote, we examine the behavior of the function as $ x $ approaches positive or negative infinity.

As $ x $ approaches positive infinity, both the numerator and the denominator grow without bound. Therefore, we can apply L'Hôpital's Rule:

$$ \lim_{x \to \infty} \frac{3e^x}{2 - 2e^x} = \lim_{x \to \infty} \frac{3e^x}{-2e^x} = -\frac{3}{2} $$

So, the horizontal asymptote is $ y = -\frac{3}{2} $.

For the slant asymptote, we can use polynomial long division to divide the numerator by the denominator. However, in this case, the degree of the numerator (1) is less than the degree of the denominator (0). Therefore, there is no slant asymptote for this function.