Question

the function f and g are defined by $f(x)=x^2$ and $g(x)=2x$, respectively. which equation is equivalent to $h(x)=\frac{f(2x)g(-2x)}{2}$? a. $h(x)=x^2 - 2x$ b. $h(x)=-8x^3$ c. $h(x)=-2x^3$ d. $h(x)=2x^3 + 2x$

Explanation:

Step1: Find \( f(2x) \)

Given \( f(x) = x^2 \), substitute \( x = 2x \) into \( f(x) \). So \( f(2x)=(2x)^2 = 4x^2 \).

Step2: Find \( g(-2x) \)

Given \( g(x)=2x \), substitute \( x = -2x \) into \( g(x) \). So \( g(-2x)=2\times(-2x)= - 4x \).

Step3: Substitute into \( h(x) \)

\( h(x)=\frac{f(2x)g(-2x)}{2}=\frac{4x^2\times(-4x)}{2} \). First, multiply the numerator: \( 4x^2\times(-4x)=-16x^3 \). Then divide by 2: \( \frac{-16x^3}{2}=-8x^3 \).

Answer:

B. \( h(x)=-8x^3 \)