Question

Using the three steps, determine the value of A such that f(x) is continuous everywhere.

f(x) = x - 3 x < 2
-1 x = 2
x^2 + Ax-5 x > 2

Answer 1

Answer:

a=0

Step-by-step explanation:

When x=2, it is at -1, so the left and right side of x=2 must approach -1, in order for it to be defined, and must be undefined at x=1 to be continuous.

We plug in 2 in the bottom equation and set it equal to -1.

$( {2}^{2} + a(2) - 5 = - 1$

$4 + 2a - 5 = - 1$

$2a = 0$

$a = 0$

So

A=0