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 $f(x)$ using the power rule $\frac{d}{dx}(ax + b)=a$ is $f^\prime(x)=2$. The left - hand derivative as $x\to - 1$ is $\lim_{x\to - 1^{-}}f^\prime(x)=2$.

Step2: Find right - hand derivative

For $x\geq - 1$, $f(x)=-x^{2}+6$. Using the power rule $\frac{d}{dx}(ax^{n})=nax^{n - 1}$, we have $f^\prime(x)=-2x$. The right - hand derivative as $x\to - 1$ is $\lim_{x\to - 1^{+}}f^\prime(x)=-2\times(-1) = 2$.

Step3: Check if derivative exists

Since $\lim_{x\to - 1^{-}}f^\prime(x)=\lim_{x\to - 1^{+}}f^\prime(x)=2$, the derivative $f^\prime(-1)$ exists and is equal to 2.

Answer:

B. 2