Question

Find b that makes the function continuous.

g(x) = (x ^ 2 - 4)/(x - 2) x < 2; (b ^ 2 - b) * x - 8 x >= 2
A) b = - 3 , b = 2
B) b = 2 , b = 4
C) b = 3 , b = - 2
D) b = 4 , b = 0​

Answer 1

Answer: C) b = 3 , b = - 2

Step-by-step explanation:

We have this piecewise function.

`g(x) = \begin{cases}(x ^ 2 - 4)/(x - 2) \ \text{ if } \ x < 2\\\\(b ^ 2 - b) * x - 8 \ \text{ if } \ x \ge 2\end{cases}`

Break each piece into a separate function.

`h(x) = (x ^ 2 - 4)/(x - 2)\\\\j(x) = (b ^ 2 - b) * x - 8`

This means g(x) = h(x) when x < 2, or g(x) = j(x) when x ≥ 2.

Let's plug x = 2 into h(x). But first we need to simplify it.

`h(x) = (x ^ 2 - 4)/(x - 2)\\\\h(x) = ((x-2)(x+2))/(x - 2)\\\\h(x) = x+2\\\\h(2) = 2+2\\\\h(2) = 4\\\\`

Then plug x = 2 into j(x).

`j(x) = (b ^ 2 - b) * x - 8\\\\j(2) = (b ^ 2 - b) * 2 - 8\\\\j(2) = 2b ^ 2 - 2b - 8\\\\`

For g(x) to be continuous at the junction point x = 2, we need to have h(2) = j(2) be true.

So,

`h(2) = j(2)\\\\4 = 2b ^ 2 - 2b - 8\\\\2b ^ 2 - 2b - 8 = 4\\\\2b ^ 2 - 2b - 8-4 = 0\\\\2b ^ 2 - 2b - 12 = 0\\\\2(b ^ 2 - b - 6) = 0\\\\2(b-3)(b+2) = 0\\\\b-3 = 0 \text{ or } b+2 = 0\\\\b = 3 \text{ or } b = -2\\\\`