Question

1. the graphs of f and g are shown to the right. if h(x)=f(x)·g(x), then h(2)=

Explanation:

Step1: Use the product rule

The product rule states that if \(h(x)=f(x)\cdot g(x)\), then \(h^{\prime}(x)=f^{\prime}(x)g(x)+f(x)g^{\prime}(x)\). So, \(h^{\prime}(2)=f^{\prime}(2)g(2)+f(2)g^{\prime}(2)\).

Step2: Find \(f(2)\) and \(g(2)\) from the graph

From the graph, \(f(2) = 1\) and \(g(2)=2\).

Step3: Find \(f^{\prime}(2)\) and \(g^{\prime}(2)\) (slopes of the tangent lines at \(x = 2\))

For \(y = f(x)\), using the slope formula \(m=\frac{y_2 - y_1}{x_2 - x_1}\). If we assume two points on \(f(x)\) (say \((-1,4)\) and \((2,1)\)), \(f^{\prime}(x)=\frac{1 - 4}{2+1}=-1\), so \(f^{\prime}(2)=-1\).
For \(y = g(x)\), if we assume two points on \(g(x)\) (say \((-3,0)\) and \((2,2)\)), \(g^{\prime}(x)=\frac{2-0}{2 + 3}=\frac{2}{5}\). But wait, another way:
The equation of \(g(x)\): Using two - point form \(y - y_1=m(x - x_1)\). Let \((x_1,y_1)=(-3,0)\) and \((x_2,y_2)=(2,2)\), \(m=\frac{2-0}{2+3}=\frac{2}{5}\), \(y=\frac{2}{5}(x + 3)\), \(g^{\prime}(x)=\frac{2}{5}\).
The equation of \(f(x)\): Let two points \((-1,4)\) and \((2,1)\), \(m=\frac{1 - 4}{2+1}=-1\), \(y-4=-1(x + 1)\), \(y=-x+3\), \(f^{\prime}(x)=-1\).
Now substitute into \(h^{\prime}(2)=f^{\prime}(2)g(2)+f(2)g^{\prime}(2)\):
\(h^{\prime}(2)=(-1)\times2+1\times\frac{2}{5}\)
\(h^{\prime}(2)=-2+\frac{2}{5}=\frac{-10 + 2}{5}=-\frac{8}{5}\).

Wait, maybe mis - read the graph.
Another approach:
The slope of \(f(x)\): \(f(x)\) is a line. If we take two points \((-1,4)\) and \((2,1)\), slope \(m_f=\frac{1 - 4}{2+1}=-1\), \(f^{\prime}(x)=-1\) for all \(x\), so \(f^{\prime}(2)=-1\), \(f(2)=1\).
The slope of \(g(x)\): take two points \((-3,0)\) and \((2,2)\), slope \(m_g=\frac{2-0}{2 + 3}=\frac{2}{5}\), \(g^{\prime}(x)=\frac{2}{5}\) for all \(x\), \(g(2)=2\).
\(h^{\prime}(2)=(-1)\times2+1\times\frac{2}{5}=-2+\frac{2}{5}=-\frac{10 - 2}{5}=-\frac{8}{5}\).
Wait, no! Wait, re - check the graph:
If \(g(x)\): when \(x=-3,y = 0\); \(x = 2,y=2\), \(g^{\prime}(x)=\frac{2-0}{2+3}=\frac{2}{5}\), \(g(2)=2\)
If \(f(x)\): when \(x=-1,y = 4\); \(x = 2,y=1\), \(f^{\prime}(x)=\frac{1 - 4}{2+1}=-1\), \(f(2)=1\)
\(h^{\prime}(2)=(-1)\times2+1\times\frac{2}{5}=-2+\frac{2}{5}=-\frac{10-2}{5}=-\frac{8}{5}\).
Wait, another thought:
The equation of \(g(x)\): \(y=\frac{2}{5}x+\frac{6}{5}\), \(g^{\prime}(x)=\frac{2}{5}\)
The equation of \(f(x)\): \(y=-x + 3\), \(f^{\prime}(x)=-1\)
\(h^{\prime}(x)=f^{\prime}(x)g(x)+f(x)g^{\prime}(x)=(-1)(\frac{2}{5}x+\frac{6}{5})+(-x + 3)\times\frac{2}{5}\)
\(h^{\prime}(x)=-\frac{2}{5}x-\frac{6}{5}-\frac{2}{5}x+\frac{6}{5}=-\frac{4}{5}x\)
\(h^{\prime}(2)=-\frac{8}{5}\)

Answer:

\(-\frac{8}{5}\)