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

[ \begin{align*} f^\prime(x)&=(0.3\times0.2x-0.3\times2)+(0.3x\times0.2 + 9\times0.2)\ &=0.06x-0.6 + 0.06x+1.8\ &=(0.06x+0.06x)+(1.8 - 0.6)\ &=0.12x + 1.2 \end{align*} ]

Answer:

$0.12x + 1.2$