Question

use the chart to the right to find h(4) for problems #23 - 28.

f(4)	f(4)	g(4)	g(4)
-8	3	3π	4

21.) h(x)=5f(x)-\frac{2}{3}g(x)
22.) h(x)=3 + 8f(x)
23.) h(x)=f(x)g(x)
24.) h(x)=\frac{f(x)}{g(x)}
25.) h(x)=\frac{g(x)}{f(x)}
28.) h(x)=\frac{f(x)+2}{-3g(x)}

Explanation:

1. Problem 21:

• # Explanation:
• ## Step1: Use the sum - difference rule of derivatives
• The sum - difference rule states that if \(h(x)=af(x)-bg(x)\), then \(h'(x)=af'(x)-bg'(x)\). Here \(a = 5\), \(b=\frac{2}{3}\), so \(h'(x)=5f'(x)-\frac{2}{3}g'(x)\).
• ## Step2: Substitute \(x = 4\)
• Given \(f'(4)=3\) and \(g'(4)=4\), we substitute these values into \(h'(x)\): \(h'(4)=5\times3-\frac{2}{3}\times4=15-\frac{8}{3}=\frac{45 - 8}{3}=\frac{37}{3}\).
• # Answer:
• \(\frac{37}{3}\)

1. Problem 22:

• # Explanation:
• ## Step1: Use the sum rule and constant - multiple rule of derivatives
• The derivative of a constant \(C\) is \(0\), and if \(h(x)=C + af(x)\), then \(h'(x)=af'(x)\). Here \(C = 3\) and \(a = 8\), so \(h'(x)=8f'(x)\).
• ## Step2: Substitute \(x = 4\)
• Since \(f'(4)=3\), then \(h'(4)=8\times3=24\).
• # Answer:
• \(24\)

1. Problem 23:

• # Explanation:
• ## Step1: Use the product rule of derivatives
• The product rule states that if \(h(x)=f(x)g(x)\), then \(h'(x)=f'(x)g(x)+f(x)g'(x)\).
• ## Step2: Substitute \(x = 4\)
• Given \(f(4)=-8\), \(f'(4)=3\), \(g(4)=3\pi\), and \(g'(4)=4\). Then \(h'(4)=3\times3\pi+( - 8)\times4=9\pi - 32\).
• # Answer:
• \(9\pi - 32\)

1. Problem 24:

• # Explanation:
• ## Step1: Use the quotient rule of derivatives
• The quotient rule states that if \(h(x)=\frac{f(x)}{g(x)}\), then \(h'(x)=\frac{f'(x)g(x)-f(x)g'(x)}{g(x)^2}\).
• ## Step2: Substitute \(x = 4\)
• Substitute \(f(4)=-8\), \(f'(4)=3\), \(g(4)=3\pi\), and \(g'(4)=4\) into the formula: \(h'(4)=\frac{3\times3\pi-(-8)\times4}{(3\pi)^2}=\frac{9\pi + 32}{9\pi^{2}}\).
• # Answer:
• \(\frac{9\pi + 32}{9\pi^{2}}\)

1. Problem 25:

• # Explanation:
• ## Step1: Use the quotient rule of derivatives
• If \(h(x)=\frac{g(x)}{f(x)}\), then \(h'(x)=\frac{g'(x)f(x)-g(x)f'(x)}{f(x)^2}\).
• ## Step2: Substitute \(x = 4\)
• Substitute \(f(4)=-8\), \(f'(4)=3\), \(g(4)=3\pi\), and \(g'(4)=4\) into the formula: \(h'(4)=\frac{4\times(-8)-3\pi\times3}{(-8)^2}=\frac{-32 - 9\pi}{64}\).
• # Answer:
• \(\frac{-32 - 9\pi}{64}\)

1. Problem 28:

• # Explanation:
• ## Step1: Use the quotient rule of derivatives
• First, if \(h(x)=\frac{f(x)+2}{-3g(x)}\), we can rewrite it as \(h(x)=-\frac{1}{3}\times\frac{f(x)+2}{g(x)}\). Using the quotient rule \((u/v)'=\frac{u'v - uv'}{v^{2}}\) where \(u = f(x)+2\), \(u'=f'(x)\), and \(v = g(x)\), \(v'=g'(x)\). So \(h'(x)=-\frac{1}{3}\times\frac{f'(x)g(x)-(f(x)+2)g'(x)}{g(x)^2}\).
• ## Step2: Substitute \(x = 4\)
• Substitute \(f(4)=-8\), \(f'(4)=3\), \(g(4)=3\pi\), and \(g'(4)=4\) into the formula:
• \(h'(4)=-\frac{1}{3}\times\frac{3\times3\pi-(-8 + 2)\times4}{(3\pi)^2}=-\frac{1}{3}\times\frac{9\pi+24}{9\pi^{2}}=-\frac{9\pi + 24}{27\pi^{2}}\).
• # Answer:
• \(-\frac{9\pi + 24}{27\pi^{2}}\)

Answer:

1. Problem 21:

• # Explanation:
• ## Step1: Use the sum - difference rule of derivatives
• The sum - difference rule states that if \(h(x)=af(x)-bg(x)\), then \(h'(x)=af'(x)-bg'(x)\). Here \(a = 5\), \(b=\frac{2}{3}\), so \(h'(x)=5f'(x)-\frac{2}{3}g'(x)\).
• ## Step2: Substitute \(x = 4\)
• Given \(f'(4)=3\) and \(g'(4)=4\), we substitute these values into \(h'(x)\): \(h'(4)=5\times3-\frac{2}{3}\times4=15-\frac{8}{3}=\frac{45 - 8}{3}=\frac{37}{3}\).
• # Answer:
• \(\frac{37}{3}\)

1. Problem 22:

• # Explanation:
• ## Step1: Use the sum rule and constant - multiple rule of derivatives
• The derivative of a constant \(C\) is \(0\), and if \(h(x)=C + af(x)\), then \(h'(x)=af'(x)\). Here \(C = 3\) and \(a = 8\), so \(h'(x)=8f'(x)\).
• ## Step2: Substitute \(x = 4\)
• Since \(f'(4)=3\), then \(h'(4)=8\times3=24\).
• # Answer:
• \(24\)

1. Problem 23:

• # Explanation:
• ## Step1: Use the product rule of derivatives
• The product rule states that if \(h(x)=f(x)g(x)\), then \(h'(x)=f'(x)g(x)+f(x)g'(x)\).
• ## Step2: Substitute \(x = 4\)
• Given \(f(4)=-8\), \(f'(4)=3\), \(g(4)=3\pi\), and \(g'(4)=4\). Then \(h'(4)=3\times3\pi+( - 8)\times4=9\pi - 32\).
• # Answer:
• \(9\pi - 32\)

1. Problem 24:

• # Explanation:
• ## Step1: Use the quotient rule of derivatives
• The quotient rule states that if \(h(x)=\frac{f(x)}{g(x)}\), then \(h'(x)=\frac{f'(x)g(x)-f(x)g'(x)}{g(x)^2}\).
• ## Step2: Substitute \(x = 4\)
• Substitute \(f(4)=-8\), \(f'(4)=3\), \(g(4)=3\pi\), and \(g'(4)=4\) into the formula: \(h'(4)=\frac{3\times3\pi-(-8)\times4}{(3\pi)^2}=\frac{9\pi + 32}{9\pi^{2}}\).
• # Answer:
• \(\frac{9\pi + 32}{9\pi^{2}}\)

1. Problem 25:

• # Explanation:
• ## Step1: Use the quotient rule of derivatives
• If \(h(x)=\frac{g(x)}{f(x)}\), then \(h'(x)=\frac{g'(x)f(x)-g(x)f'(x)}{f(x)^2}\).
• ## Step2: Substitute \(x = 4\)
• Substitute \(f(4)=-8\), \(f'(4)=3\), \(g(4)=3\pi\), and \(g'(4)=4\) into the formula: \(h'(4)=\frac{4\times(-8)-3\pi\times3}{(-8)^2}=\frac{-32 - 9\pi}{64}\).
• # Answer:
• \(\frac{-32 - 9\pi}{64}\)

1. Problem 28:

• # Explanation:
• ## Step1: Use the quotient rule of derivatives
• First, if \(h(x)=\frac{f(x)+2}{-3g(x)}\), we can rewrite it as \(h(x)=-\frac{1}{3}\times\frac{f(x)+2}{g(x)}\). Using the quotient rule \((u/v)'=\frac{u'v - uv'}{v^{2}}\) where \(u = f(x)+2\), \(u'=f'(x)\), and \(v = g(x)\), \(v'=g'(x)\). So \(h'(x)=-\frac{1}{3}\times\frac{f'(x)g(x)-(f(x)+2)g'(x)}{g(x)^2}\).
• ## Step2: Substitute \(x = 4\)
• Substitute \(f(4)=-8\), \(f'(4)=3\), \(g(4)=3\pi\), and \(g'(4)=4\) into the formula:
• \(h'(4)=-\frac{1}{3}\times\frac{3\times3\pi-(-8 + 2)\times4}{(3\pi)^2}=-\frac{1}{3}\times\frac{9\pi+24}{9\pi^{2}}=-\frac{9\pi + 24}{27\pi^{2}}\).
• # Answer:
• \(-\frac{9\pi + 24}{27\pi^{2}}\)