Question

determine if the following piece - wise defined function is differentiable at x = 0.
f(x)=\begin{cases}3x - 5, & xgeq0\\x^{2}+4x - 5, & x < 0end{cases}
what is the right - hand derivative of the given function?
(lim_{h
ightarrow0^{+}}\frac{f(0 + h)-f(0)}{h}=3) (type an integer or a simplified fraction.)
what is the left - hand derivative of the given function?
(lim_{h
ightarrow0^{-}}\frac{f(0 + h)-f(0)}{h}=square) (type an integer or a simplified fraction.)

Explanation:

Step1: Find \(f(0)\)

For \(x\geq0\), \(f(x)=3x - 5\), so \(f(0)=3\times0 - 5=- 5\).

Step2: Calculate the left - hand derivative

For \(x<0\), \(f(x)=x^{2}+4x - 5\). Then \(f(0 + h)=h^{2}+4h - 5\) (since \(h\to0^{-}\)).
The left - hand derivative is \(\lim_{h\to0^{-}}\frac{f(0 + h)-f(0)}{h}=\lim_{h\to0^{-}}\frac{h^{2}+4h - 5-(-5)}{h}\).

Step3: Simplify the limit

\(\lim_{h\to0^{-}}\frac{h^{2}+4h - 5 + 5}{h}=\lim_{h\to0^{-}}\frac{h^{2}+4h}{h}=\lim_{h\to0^{-}}(h + 4)\).
As \(h\to0^{-}\), \(\lim_{h\to0^{-}}(h + 4)=4\).

Answer:

4