if $f(x)=4 + 7x-2x^{2}$, find $f(-3)$.

Explanation:

Step1: Find the derivative of $f(x)$

Using the power - rule $\frac{d}{dx}(x^n)=nx^{n - 1}$, for $f(x)=4 + 7x-2x^{2}$, we have $f'(x)=\frac{d}{dx}(4)+\frac{d}{dx}(7x)-\frac{d}{dx}(2x^{2})$. Since $\frac{d}{dx}(c)=0$ (where $c$ is a constant), $\frac{d}{dx}(7x)=7$ and $\frac{d}{dx}(2x^{2}) = 4x$. So $f'(x)=0 + 7-4x=7 - 4x$.

Step2: Evaluate $f'(-3)$

Substitute $x=-3$ into $f'(x)$. We get $f'(-3)=7-4\times(-3)$. $f'(-3)=7 + 12$.

Answer:

$19$