Question

f(x) = \

$$\begin{cases}2x + 5&\\text{for }x < - 1\\-x^{2}+6&\\text{for }x\\geq - 1\\end{cases}$$

if f is the function defined above, then f(-1) is
a -2
b 2
c 3
d 5
e nonexistent

Explanation:

Step1: Find left - hand derivative

For \(x < - 1\), \(f(x)=2x + 5\). The derivative of \(y = 2x+5\) using the power rule \((x^n)^\prime=nx^{n - 1}\) is \(f^\prime(x)=2\). The left - hand derivative as \(x\to - 1\) is \(2\).

Step2: Find right - hand derivative

For \(x\geq - 1\), \(f(x)=-x^{2}+6\). Using the power rule, \(f^\prime(x)=-2x\). The right - hand derivative as \(x\to - 1\) is \(f^\prime(-1)=-2\times(-1) = 2\).

Step3: Check differentiability

Since the left - hand derivative \(=2\) and the right - hand derivative \(=2\), the derivative \(f^\prime(-1)\) exists and \(f^\prime(-1)=2\).

Answer:

B. 2