Question

question let h(x)=f(x)g(x). if f(x)=-2x^2 + 2x + 3 and g(x)=-x^2 - 3x - 4, what is h(-1)? provide your answer below: h(-1)=□

Explanation:

Step1: Apply the product - rule

The product - rule states that if $h(x)=f(x)g(x)$, then $h^{\prime}(x)=f^{\prime}(x)g(x)+f(x)g^{\prime}(x)$. First, find the derivatives of $f(x)$ and $g(x)$.
For $f(x)=-2x^{2}+2x + 3$, using the power - rule $\frac{d}{dx}(ax^{n})=nax^{n - 1}$, we have $f^{\prime}(x)=\frac{d}{dx}(-2x^{2}+2x + 3)=-4x + 2$.
For $g(x)=-x^{2}-3x - 4$, using the power - rule, we have $g^{\prime}(x)=\frac{d}{dx}(-x^{2}-3x - 4)=-2x-3$.

Step2: Evaluate $f(x)$, $f^{\prime}(x)$, $g(x)$, and $g^{\prime}(x)$ at $x=-1$

Evaluate $f(-1)$:
$f(-1)=-2(-1)^{2}+2(-1)+3=-2 - 2 + 3=-1$.
Evaluate $f^{\prime}(-1)$:
$f^{\prime}(-1)=-4(-1)+2=4 + 2=6$.
Evaluate $g(-1)$:
$g(-1)=-(-1)^{2}-3(-1)-4=-1 + 3-4=-2$.
Evaluate $g^{\prime}(-1)$:
$g^{\prime}(-1)=-2(-1)-3=2 - 3=-1$.

Step3: Calculate $h^{\prime}(-1)$

Substitute the values into the product - rule formula $h^{\prime}(-1)=f^{\prime}(-1)g(-1)+f(-1)g^{\prime}(-1)$.
$h^{\prime}(-1)=(6)\times(-2)+(-1)\times(-1)=-12 + 1=-11$.

Answer:

$-11$