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

Question

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

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

a) If f is continuous at -3, then the graph cannot jump at that point.

(2x² + 5x -3)/(x²+ 4x +3) must = (kx + 1/2) when x= -3 by substitution you get

(18 - 15 -3)/(9 - 12 +3) = -3k + 1/2

0 = -3k + 1/2

-1/2 = -3k

k = - 1/6

b) at x=0 There is a HA (on the right) in the graph because the denominator in the last function = 0 which is undefined.

c) horizontal Asymptotes (HA) appear when the denominator = 0

The first part of the function the denominator is x²+ 4x -3 =0 the solution is -2, but that isn't part of the domain (x<-3), so no HA there.

The second part the only denominator is 2 and 2 cannot =0, no HA

The third part, 3^x - 1 =0 when x=0 there is a HA as x –> 0⁺