Question

determine if the piece - wise - defined function is differentiable at x = 0.
f(x)=
\

$$\begin{cases}2x+\tan x, & xgeq0 \\\\8x^{2}, & x < 0\\end{cases}$$

select the correct choice below and, if necessary, fill in the answer boxes in your choice.
\\(\bigcirc\\)a. the function is not differentiable at x = 0 because \\(\lim_{h\to0^{-}}\frac{f(0 + h)-f(0)}{h}=\square\\) and \\(\lim_{h\to0^{+}}\frac{f(0 + h)-f(0)}{h}=\square\\).
(type integers or simplified fractions.)
\\(\bigcirc\\)b. the function is differentiable at x = 0 because it is continuous at x = 0 and \\(\lim_{h\to0^{-}}\frac{f(0 + h)-f(0)}{h}=\square\\) and \\(\lim_{h\to0^{+}}\frac{f(0 + h)-f(0)}{h}=\square\\).
(type integers or simplified fractions.)
\\(\bigcirc\\)c. the function is not differentiable at x = 0 because it is not continuous at x = 0.

Explanation:

Step1: Find the left - hand limit of the difference quotient

For \(x<0\), \(f(x) = 8x^{2}\), and \(f(0)=2\times0+\tan(0) = 0\).
The left - hand limit as \(h\to0^{-}\) of \(\frac{f(0 + h)-f(0)}{h}=\lim_{h\to0^{-}}\frac{8h^{2}-0}{h}=\lim_{h\to0^{-}}8h = 0\)

Step2: Find the right - hand limit of the difference quotient

For \(x\geq0\), \(f(x)=2x+\tan x\).
The right - hand limit as \(h\to0^{+}\) of \(\frac{f(0 + h)-f(0)}{h}=\lim_{h\to0^{+}}\frac{2h+\tan h - 0}{h}=\lim_{h\to0^{+}}(2+\frac{\tan h}{h})\)
Since \(\lim_{h\to0^{+}}\frac{\tan h}{h}=1\), then \(\lim_{h\to0^{+}}(2+\frac{\tan h}{h})=2 + 1=3\)

Since \(\lim_{h\to0^{-}}\frac{f(0 + h)-f(0)}{h}=0\) and \(\lim_{h\to0^{+}}\frac{f(0 + h)-f(0)}{h}=3\), the left - hand and right - hand limits of the difference quotient are not equal.

Answer:

A. The function is not differentiable at \(x = 0\) because \(\lim_{h\to0^{-}}\frac{f(0 + h)-f(0)}{h}=0\) and \(\lim_{h\to0^{+}}\frac{f(0 + h)-f(0)}{h}=3\)