Question

Need help finding a constant k at which the function is continuous everywhere.

Answer 1

Answer: 36k = k - 6, k = -6/35

Explanation:

The two parts are continuous by themselves. If either weren't, no value of k could fix it.

The only point of discontinuity is x=-6. The function will be continuous for any and all values of k which make k(-6)^2 = -6+k. 36k = k-6, 35k=-6, k=-6/35.

Answer 2

k = -6/35

Explanation:

To make the function continuous

kx^2 = x+k

These must be equal where the function is defined for two different intervals

This is at the point x=-6 so let x=-6

k(-6)^2 = -6+k

36k = -6+k

Subtract k from each side

36k-k = -6+k-k

35k = -6

Divide by 35

35k/35 = -6/35

k = -6/35