Question

1. let ( h(x)=f(x)+g(x) ). find a. ( h^{prime}(1) ), b. ( h^{prime}(3) ), and c. ( h^{prime}(4) ).

Explanation:

Step1: Find the derivative of \(f(x)\) and \(g(x)\)

The derivative of a linear function \(y = mx + b\) is its slope \(m\).
For \(f(x)\):
The line \(f(x)\) passes through \((0,4)\) and \((2,2)\). The slope \(m_f=\frac{2 - 4}{2-0}=- 1\), so \(f^{\prime}(x)=-1\) for all \(x\) in its domain.
For \(g(x)\):
For \(x<3\), \(g(x)\) is a line passing through \((0,0)\) and \((2,2)\). The slope \(m_{g1}=\frac{2 - 0}{2-0}=1\), so \(g^{\prime}(x)=1\) for \(x < 3\).
For \(x>3\), \(g(x)\) is a line. Let's find its slope. Suppose two points \((3,1)\) and \((5,3)\). The slope \(m_{g2}=\frac{3 - 1}{5 - 3}=1\).

Step2: Use the sum - rule of derivatives

The sum - rule states that if \(h(x)=f(x)+g(x)\), then \(h^{\prime}(x)=f^{\prime}(x)+g^{\prime}(x)\)

Part a: Find \(h^{\prime}(1)\)

Since \(1<3\), \(g^{\prime}(1) = 1\) and \(f^{\prime}(1)=-1\)
\(h^{\prime}(1)=f^{\prime}(1)+g^{\prime}(1)\)
\(h^{\prime}(1)=-1 + 1\)
\(h^{\prime}(1)=0\)

Part b: Find \(h^{\prime}(3)\)

The function \(g(x)\) has a corner at \(x = 3\), but \(f(x)\) is differentiable everywhere. However, since \(f^{\prime}(x)=-1\) and \(g(x)\) has a non - differentiable point at \(x = 3\). But if we consider the left - hand and right - hand derivatives:
Left - hand derivative of \(h(x)\) at \(x = 3\): \(h_{-}^{\prime}(3)=f^{\prime}(3)+g_{-}^{\prime}(3)\), \(f^{\prime}(3)=-1\), \(g_{-}^{\prime}(3)=1\), so \(h_{-}^{\prime}(3)=-1 + 1=0\)
Right - hand derivative of \(h(x)\) at \(x = 3\): \(h_{+}^{\prime}(3)=f^{\prime}(3)+g_{+}^{\prime}(3)\), \(f^{\prime}(3)=-1\), \(g_{+}^{\prime}(3)=1\), so \(h_{+}^{\prime}(3)=-1 + 1=0\)
\(h^{\prime}(3)=0\)

Part c: Find \(h^{\prime}(4)\)

Since \(4>3\), \(g^{\prime}(4) = 1\) and \(f^{\prime}(4)=-1\)
\(h^{\prime}(4)=f^{\prime}(4)+g^{\prime}(4)\)
\(h^{\prime}(4)=-1+1\)
\(h^{\prime}(4)=0\)

Answer:

a. \(h^{\prime}(1)=0\)
b. \(h^{\prime}(3)=0\)
c. \(h^{\prime}(4)=0\)