Question

1. $f(x)=4 - x^{2}$; $f(-3), f(0), f(1)$

Explanation:

Step1: Differentiate the function

Using the power - rule, if $f(x)=4 - x^{2}$, then $f'(x)=\frac{d}{dx}(4)-\frac{d}{dx}(x^{2})$. Since $\frac{d}{dx}(c) = 0$ for a constant $c$ and $\frac{d}{dx}(x^{n})=nx^{n - 1}$, we have $f'(x)=0-2x=-2x$.

Step2: Calculate $f'(-3)$

Substitute $x = - 3$ into $f'(x)$. So $f'(-3)=-2\times(-3)=6$.

Step3: Calculate $f'(0)$

Substitute $x = 0$ into $f'(x)$. So $f'(0)=-2\times0 = 0$.

Step4: Calculate $f'(1)$

Substitute $x = 1$ into $f'(x)$. So $f'(1)=-2\times1=-2$.

Answer:

$f'(-3)=6$, $f'(0)=0$, $f'(1)=-2$