Find value of k if the funtion is to be continuous at x=1?

Question

f(x) = both ((x^2-4 if x<1) and (kx+2 if x>or equal to 1))

thanks

Answer 1

For a piecewise function to be continuous at x = 1, the functions must have the same value at x = 1. Therefore, we'll just plug in 1 for x into the equations and set them equal to each other, and we can solve for k:

x^2 - 4 = kx + 2
==> plug in x = 1
1^2 - 4 = k(1) + 2
==> simplify
-3 = k + 2
==> subtract 2 from both sides
k = -5

Let's check: 1^2 - 4 = -5 + 2 = -3 ... IT WORKS

ANSWER: k = -5.

Answer 2

To be continuous the functions must have the same value at x = 1:

x^2 - 4 = kx + 2
1 - 4 = k + 2 or k = -5

Answer 3

basically you want
x^2 - 4 = kx + 2
at x = 1

So that the piecewise function will meet at x=1

(1)^2 - 4 = k + 2
-3 = k + 2
k = -1