Question

question let h(x)=f(x)g(x). if f(x)=3x^2 - x - 4 and g(x)=-x^2 + 2, what is h(-1)? do not include \h(-1)=\ in your answer. for example, if you found h(-1)=7, you would enter 7. provide your answer below:

Explanation:

Step1: Find derivatives of $f(x)$ and $g(x)$

$f'(x)=\frac{d}{dx}(3x^{2}-x - 4)=6x - 1$; $g'(x)=\frac{d}{dx}(-x^{2}+2)=-2x$

Step2: Apply the product - rule

The product - rule states that if $h(x)=f(x)g(x)$, then $h'(x)=f'(x)g(x)+f(x)g'(x)$.

Step3: Substitute $x = - 1$ into $f(x)$, $f'(x)$, $g(x)$ and $g'(x)$

$f(-1)=3\times(-1)^{2}-(-1)-4=3 + 1-4=0$; $f'(-1)=6\times(-1)-1=-6 - 1=-7$; $g(-1)=-(-1)^{2}+2=-1 + 2=1$; $g'(-1)=-2\times(-1)=2$

Step4: Calculate $h'(-1)$

$h'(-1)=f'(-1)g(-1)+f(-1)g'(-1)=(-7)\times1+0\times2=-7$

Answer:

-7