Let f be the function defined below, where k is a constant

Question

f(x)= ( 2x^2+5x-3/x^2+4x+3 for x<-3

<(kx+1/2 for -3<x<0

(2^x/3^x-1 for x>0

A. For what value of k, if any, is continuous at x=-3? Justify your answer. B. What type of discontinuity does f have at x=0? Give a reason for your answer. C. Find all horizontal asymptotes to the graph of f. Show the work that leads to your answer.

P.S.- The 3 parentheses on the way left of the function is one big left parenthesis. It's a special one though with a point sticking to the left so I used a less than sign to represent it.

Answer 1

Let me first rewrite the function definition the way I understand it, including parentheses for disambiguation.

If I guessed wrong, hopefully you can still apply the approach to your correct function.

f(x) = (2x² + 5x - 3)/(x² + 4x +3) for x < -3

= kx + 1/2 for -3 ≤ x ≤ 0

= 2^x/(3^x - 1) for x > 0

A.

First calculate the limit as x -> -3 from the left.

As you can see the expression (2x² + 5x - 3)/(x² + 4x + 3)

would have 0 in both numerator and denominator if we were to set x=-3.

Therefore we can apply the L'Hopital Rule.

lim (2x² + 5x - 3)/(x² + 4x +3) =

lim (4x + 5)/(2x + 4) =

lim (4×(-3) + 5)/(2×(-3) + 4) = 7/2

To make the function continuous at -3, we need

f(-3) = 7/2

-3k + 1/2 = 7/2

k = -1

This setting of k makes the limit on the left equal to f(-3).

The limit on the right is automatically equal f(-3) because (kx + 1/2) is continuous for any k.

B.

Consider the limit of f(x) when x->0 from the right.

If we were to plug x=0 into 2^x/(3^x - 1) we would get 1 in the numerator and 0 in the denominator.

Therefore the limit is ∞.

The value f(0) = 1/2.

Therefore the function is not continuous at 0.

If we are using the same terminology, that I would classify the discontinuity as infinite.

C.

Horizontal asymptote on the left is the limit of f(x) as x->-∞.

lim f(x) = lim (2x² + 5x - 3)/(x² + 4x + 3) = lim 2x²/x² = lim 2 = 2.

Horizontal asymptote on the right is the limit of f(x) as x->∞.

lim f(x) = 2^x/(3^x - 1) = lim 2^x/3^x = lim (2/3)^x = 0