Question

1. determine all values at which the following functions are not differentiable, if any exist. justify your answers.

(a) $f(x)=x|x|$
(b) $g(x)=\begin{cases}\frac{1}{x + 2}&\text{if }x < - 1\\3x&\text{if }-1leq xleq3\\|x + 3|+3&\text{if }x>3end{cases}$

Explanation:

Subsection (a)

Step1: Rewrite the function

Rewrite \(f(x)=x|x|\) as a piece - wise function. When \(x\geq0\), \(|x| = x\) and \(f(x)=x\cdot x=x^{2}\); when \(x<0\), \(|x|=-x\) and \(f(x)=x\cdot(-x)=-x^{2}\).

Step2: Calculate the derivative for each piece

For \(y = x^{2}\), \(y^\prime=2x\) when \(x > 0\); for \(y=-x^{2}\), \(y^\prime=-2x\) when \(x<0\).
At \(x = 0\), the left - hand derivative \(\lim_{x
ightarrow0^{-}}\frac{f(x)-f(0)}{x - 0}=\lim_{x
ightarrow0^{-}}\frac{-x^{2}-0}{x}=\lim_{x
ightarrow0^{-}}(-x)=0\).
The right - hand derivative \(\lim_{x
ightarrow0^{+}}\frac{f(x)-f(0)}{x - 0}=\lim_{x
ightarrow0^{+}}\frac{x^{2}-0}{x}=\lim_{x
ightarrow0^{+}}x = 0\). So \(f(x)\) is differentiable everywhere.

Subsection (b)

Step1: Analyze the first - piece function

For \(g(x)=\frac{1}{x + 2}\), \(x<-1\). The derivative \(g^\prime(x)=-\frac{1}{(x + 2)^{2}}\), and it is differentiable for all \(x<-1\) since the function is a rational function and the denominator is non - zero in this interval.

Step2: Analyze the second - piece function

For \(g(x)=3x\), \(-1\leq x\leq3\). The derivative \(g^\prime(x)=3\), and it is differentiable for all \(x\in(-1,3)\).

Step3: Check the differentiability at \(x=-1\)

Left - hand derivative at \(x=-1\): \(\lim_{x
ightarrow - 1^{-}}g^\prime(x)=\lim_{x
ightarrow - 1^{-}}-\frac{1}{(x + 2)^{2}}=-1\).
Right - hand derivative at \(x=-1\): \(\lim_{x
ightarrow - 1^{+}}g^\prime(x)=\lim_{x
ightarrow - 1^{+}}3 = 3\). Since the left - hand and right - hand derivatives are not equal, \(g(x)\) is not differentiable at \(x=-1\).

Step4: Analyze the third - piece function

For \(g(x)=|x + 3|+3\), \(x>3\). When \(x>3\), \(g(x)=(x + 3)+3=x + 6\), and \(g^\prime(x)=1\).

Step5: Check the differentiability at \(x = 3\)

Left - hand derivative at \(x = 3\): \(\lim_{x
ightarrow3^{-}}g^\prime(x)=\lim_{x
ightarrow3^{-}}3=3\).
Right - hand derivative at \(x = 3\): \(\lim_{x
ightarrow3^{+}}g^\prime(x)=\lim_{x
ightarrow3^{+}}1 = 1\). Since the left - hand and right - hand derivatives are not equal, \(g(x)\) is not differentiable at \(x = 3\).

Answer:

(a) The function \(f(x)=x|x|\) is differentiable everywhere.
(b) The function \(g(x)\) is not differentiable at \(x=-1\) and \(x = 3\).