Question

question
let $f(x)=\begin{cases}(x + 9)^2&x < - 8\\3x + 4&-8leq x < 2\sqrt{x - 2}&xgeq2end{cases}$
which of the following statements is true?
a) $f(x)$ is increasing for all $x$.
b) $f(0)=81$
c) there is an open - circle at $(2,10)$.
d) $f(2)$ is undefined.

Explanation:

Step1: Evaluate \(f(0)\)

Since \(- 8\leq0<2\), use \(f(x)=3x + 4\). Then \(f(0)=3\times0 + 4=4
eq81\), so B is false.

Step2: Evaluate \(f(2)\)

Since \(x = 2\) and for \(x\geq2\), \(f(x)=\sqrt{x - 2}\), then \(f(2)=\sqrt{2-2}=0\), so D is false.

Step3: Analyze the continuity at \(x = 2\)

For \(x<2\), when \(x\) approaches \(2\) from the left, \(y = 3x+4\), and \(\lim_{x
ightarrow2^{-}}(3x + 4)=3\times2+4 = 10\). For \(x\geq2\), \(f(2)=\sqrt{2 - 2}=0\). There is a break - point at \(x = 2\), and there is an open - circle at the point \((2,10)\) (the left - hand limit value).

Step4: Analyze the increasing nature

For \(x<-8\), \(y=(x + 9)^2\), its derivative \(y'=2(x + 9)\), which is negative for \(x<-9\) and positive for \(-9

Answer:

C. There is an open circle at \((2,10)\)