Question

answer this question after you finish topic 2.5 (parts a and b) and after topic 2.7 (part c). 6.) for each function, (f(x)), determine if the function is continuous or non - continuous, differentiable or non - differentiable, and sketch the curve. you must show thorough work in order to justify your answer. a.) (f(x)=\begin{cases}x^{2},&xgeq0\\x,&x < 0end{cases}) b.) (f(x)=\begin{cases}4 - x^{2},&x < 1\\2x + 2,&xgeq1end{cases}) continuous differentiable both neither continuous differentiable both neither c.) (f(x)=\begin{cases}cos x,&xgeq0\\1 - x^{2},&x < 0end{cases}) note: exercise caution when graphing (f(x)=cos x) using the provided coordinate axes.

Explanation:

Step1: Check continuity at the break - point for part a

For \(f(x)=

$$\begin{cases}x^{2},&x\geq0\\x,&x < 0\end{cases}$$

\), find the left - hand limit \(\lim_{x
ightarrow0^{-}}f(x)=\lim_{x
ightarrow0^{-}}x = 0\) and the right - hand limit \(\lim_{x
ightarrow0^{+}}f(x)=\lim_{x
ightarrow0^{+}}x^{2}=0\), and \(f(0) = 0^{2}=0\). So the function is continuous at \(x = 0\). Then find the left - hand derivative \(f_{-}'(0)=\lim_{h
ightarrow0^{-}}\frac{f(0 + h)-f(0)}{h}=\lim_{h
ightarrow0^{-}}\frac{h - 0}{h}=1\) and the right - hand derivative \(f_{+}'(0)=\lim_{h
ightarrow0^{+}}\frac{f(0 + h)-f(0)}{h}=\lim_{h
ightarrow0^{+}}\frac{h^{2}-0}{h}=0\). Since \(f_{-}'(0)
eq f_{+}'(0)\), the function is non - differentiable at \(x = 0\).

Step2: Check continuity at the break - point for part b

For \(f(x)=

$$\begin{cases}4 - x^{2},&x<1\\2x + 2,&x\geq1\end{cases}$$

\), \(\lim_{x
ightarrow1^{-}}f(x)=\lim_{x
ightarrow1^{-}}(4 - x^{2})=4-1^{2}=3\), \(\lim_{x
ightarrow1^{+}}f(x)=\lim_{x
ightarrow1^{+}}(2x + 2)=2\times1+2 = 4\), and \(f(1)=2\times1 + 2=4\). Since \(\lim_{x
ightarrow1^{-}}f(x)
eq\lim_{x
ightarrow1^{+}}f(x)\), the function is non - continuous at \(x = 1\), and thus non - differentiable at \(x = 1\).

Step3: Check continuity at the break - point for part c

For \(f(x)=

$$\begin{cases}\cos x,&x\geq0\\1 - x^{2},&x<0\end{cases}$$

\), \(\lim_{x
ightarrow0^{-}}f(x)=\lim_{x
ightarrow0^{-}}(1 - x^{2})=1\), \(\lim_{x
ightarrow0^{+}}f(x)=\lim_{x
ightarrow0^{+}}\cos x=1\), and \(f(0)=\cos(0)=1\). So the function is continuous at \(x = 0\). The left - hand derivative \(f_{-}'(0)=\lim_{h
ightarrow0^{-}}\frac{f(0 + h)-f(0)}{h}=\lim_{h
ightarrow0^{-}}\frac{1 - h^{2}-1}{h}=0\), and the right - hand derivative \(f_{+}'(0)=\lim_{h
ightarrow0^{+}}\frac{f(0 + h)-f(0)}{h}=\lim_{h
ightarrow0^{+}}\frac{\cos h - 1}{h}=0\). So the function is differentiable at \(x = 0\).

Answer:

a. continuous
b. neither
c. both