f(x)=(2^x)/x^2

Question

(a) Find the domain of f(x).

(b) Find the horizontal asymptotes of f(x).

(c) Find the vertical asymptotes of f(x).

(d) Find the x intercepts. If there are none, state so.

Answer 1

f(x) = (2^x) / x²

 

 

Part a

 

The denominator of a function cannot be equal to zero.  Therefore, your domain is

 

(-∞,0)∪(0,∞)

 

 

Part b

 

The horizontal asymptote is the line where the function gets closer to, but never touches, as the x values increase or decrease infinitely from either left or right.  In order to find the horizontal asymptote, we must find the limit of f(x) as x approaches infinity.

 

lim       (2^x) / x²

x -->∞

 

Notice that when we plug in ∞, f(x) will be ∞/∞.  This is undeterminate.  So we must use L'Hosiptal Rule.  We take the derivative of the numerator divided by the derivative of the denominator.  We do this until the limit is no longer undeterminate.

 

lim = (2^xln2) / 2x

x-->∞

 

lim = (2^xln2*ln2) / 2

x-->∞

 

Based on this, there is no steady horizontal line because the limit is infinity and not a constant number.

 

 

Part c

 

Set the denominator of f(x) equal to zero. 

 

x² = 0

x = 0

 

The vertical asymptote is x = 0.

 

 

Part d

 

We set the numerator of f(x) equal to zero.

 

2^x = 0

 

There are no x-incercepts because there is no exponential value of any number that will give us a result of zero. 

Here is why.  If we substitute   -2 ≤ x ≤ -2,

 

2² = 4

2¹ = 2

2⁰ = 1

2^-1 = 1/2

2^-2 =1/4