10. state the x values of the critical points for the following absolute value functions: (2 points…

Explanation:

Step1: Recall the property of absolute value function

For an absolute value function (y = |u(x)|+k), the critical point occurs where (u(x)=0) because the derivative of (y = |u(x)|) changes its sign at (u(x) = 0).

Step2: Solve for the critical - point of (f(x))

Set (3x + 6=0). [ \begin{align*} 3x+6&=0\ 3x&=- 6\ x&=-2 \end{align*} ]

Step3: Solve for the critical - point of (g(x))

Set (\frac{x - 2}{3}=0). [ \begin{align*} \frac{x - 2}{3}&=0\ x-2&=0\ x&=2 \end{align*} ]

Answer:

For (f(x)=|3x + 6|-1), the critical point is (x=-2). For (g(x)=\left|\frac{x - 2}{3}\right|+5), the critical point is (x = 2).