Question

find f(x).
f(x)=(0.3x + 9)(0.2x - 2)
f(x)=□

Explanation:

Step1: Apply product - rule

The product - rule states that if $y = u(x)v(x)$, then $y^\prime=u^\prime(x)v(x)+u(x)v^\prime(x)$. Let $u(x)=0.3x + 9$ and $v(x)=0.2x - 2$.

Step2: Find $u^\prime(x)$ and $v^\prime(x)$

Differentiate $u(x)$ with respect to $x$: $u^\prime(x)=\frac{d}{dx}(0.3x + 9)=0.3$. Differentiate $v(x)$ with respect to $x$: $v^\prime(x)=\frac{d}{dx}(0.2x - 2)=0.2$.

Step3: Substitute into product - rule formula

$f^\prime(x)=u^\prime(x)v(x)+u(x)v^\prime(x)=0.3(0.2x - 2)+(0.3x + 9)\times0.2$.

Step4: Expand and simplify

$$
LATEXBLOCK0
$$

Answer:

$0.12x + 1.2$