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}=square \text{ (type an integer or a simplified fraction.)}

Explanation:

Step1: Find f(0)

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

Step2: Find f(0 + h) for \(h>0\)

Since \(0 + h>0\) when \(h>0\), \(f(0 + h)=3(0 + h)-5=3h-5\).

Step3: Calculate the right - hand derivative

Substitute \(f(0 + h)\) and \(f(0)\) into the right - hand derivative formula \(\lim_{h
ightarrow0^{+}}\frac{f(0 + h)-f(0)}{h}\).
\[

$$\begin{align*} \lim_{h ightarrow0^{+}}\frac{(3h - 5)-(-5)}{h}&=\lim_{h ightarrow0^{+}}\frac{3h-5 + 5}{h}\\ &=\lim_{h ightarrow0^{+}}\frac{3h}{h}\\ &=\lim_{h ightarrow0^{+}}3\\ &=3 \end{align*}$$

\]

Answer:

3