Question

determine if the following piecewise - defined function is differentiable at x = 0.
f(x)=\begin{cases}2x - 3, & xgeq0\\x^{2}+3x - 3, & x < 0end{cases}
select the correct choice below and, if necessary, fill in the answer boxes within your choice
a. the function is differentiable at x = 0 because it is continuous at x = 0 and (lim_{h
ightarrow0^{-}}\frac{f(0 + h)-f(0)}{h}=) and (lim_{h
ightarrow0^{+}}\frac{f(0 + h)-f(0)}{h}=)
(type integers or simplified fractions.)
b. the function is not differentiable at x = 0 because it is continuous at x = 0 and (lim_{h
ightarrow0^{-}}\frac{f(0 + h)-f(0)}{h}=) and (lim_{h
ightarrow0^{+}}\frac{f(0 + h)-f(0)}{h}=)
(type integers or simplified fractions.)
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 \(x<0\), \(f(x)=x^{2}+3x - 3\). First, find \(f(0)\) using the part of the function for \(x\geq0\), so \(f(0)=2\times0 - 3=-3\).
The left - hand limit of the difference quotient \(\lim_{h
ightarrow0^{-}}\frac{f(0 + h)-f(0)}{h}=\lim_{h
ightarrow0^{-}}\frac{(h^{2}+3h - 3)-(-3)}{h}=\lim_{h
ightarrow0^{-}}\frac{h^{2}+3h}{h}=\lim_{h
ightarrow0^{-}}(h + 3)=3\).

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

For \(x\geq0\), \(f(x)=2x - 3\). The right - hand limit of the difference quotient \(\lim_{h
ightarrow0^{+}}\frac{f(0 + h)-f(0)}{h}=\lim_{h
ightarrow0^{+}}\frac{(2h-3)-(-3)}{h}=\lim_{h
ightarrow0^{+}}\frac{2h}{h}=2\).
Since \(\lim_{h
ightarrow0^{-}}\frac{f(0 + h)-f(0)}{h}=3\) and \(\lim_{h
ightarrow0^{+}}\frac{f(0 + h)-f(0)}{h}=2\), the left - hand and right - hand limits of the difference quotient are not equal.

Answer:

B. The function is not differentiable at \(x = 0\) because it is continuous at \(x = 0\) and \(\lim_{h
ightarrow0^{-}}\frac{f(0 + h)-f(0)}{h}=3\) and \(\lim_{h
ightarrow0^{+}}\frac{f(0 + h)-f(0)}{h}=2\)