Question

the graphs of the piecewise linear functions f and g are shown above. if the function h is defined by h(x)=f(x)g(x), then h(2) is

Explanation:

Step1: Find the derivatives of \(f(x)\) and \(g(x)\) at \(x = 2\)

The slope of a linear function \(y=mx + b\) (where \(m\) is the slope) is used to find the derivative.
For \(y = f(x)\):
Using two points \((0,0)\) and \((6,3)\) on the non - vertical part of \(y = f(x)\) near \(x = 2\), the slope \(m_f=\frac{3-0}{6 - 0}=\frac{1}{2}\), so \(f^{\prime}(2)=\frac{1}{2}\)
For \(y = g(x)\):
Using two points \((3,0)\) and \((6,3)\) on the non - vertical part of \(y = g(x)\) near \(x = 2\), the slope \(m_g=\frac{3-0}{6 - 3}=1\), so \(g^{\prime}(2)=1\)
Also, \(f(2)=\frac{1}{2}\times2 = 1\) and \(g(2)=1\times(2 - 3)+0=- 1\)

Step2: Use the product rule

The product rule states that if \(h(x)=f(x)g(x)\), then \(h^{\prime}(x)=f^{\prime}(x)g(x)+f(x)g^{\prime}(x)\)
Substitute \(x = 2\) into the product rule formula:
\(h^{\prime}(2)=f^{\prime}(2)g(2)+f(2)g^{\prime}(2)\)
\(h^{\prime}(2)=\frac{1}{2}\times(-1)+1\times1\)
\(h^{\prime}(2)=-\frac{1}{2}+1=\frac{1}{2}\)

Answer:

A. 2