use the chart to the right to find h(4) for problems #23 - 28.\n| f(4) | f(4) | g(4) | g(4) |\n| -8 | 3 | 3π…

use the chart to the right to find h(4) for problems #23 - 28.\n| f(4) | f(4) | g(4) | g(4) |\n| -8 | 3 | 3π | 4 |\n21.) h(x)=5f(x)-\\frac{2}{3}g(x)\n22.) h(x)=3 + 8f(x)\n23.) h(x)=f(x)g(x)\n24.) h(x)=\\frac{f(x)}{g(x)}\n25.) h(x)=\\frac{g(x)}{f(x)}\n28.) h(x)=\\frac{f(x)+2}{-3g(x)}

use the chart to the right to find h(4) for problems #23 - 28.\n| f(4) | f(4) | g(4) | g(4) |\n| -8 | 3 | 3π | 4 |\n21.) h(x)=5f(x)-\\frac{2}{3}g(x)\n22.) h(x)=3 + 8f(x)\n23.) h(x)=f(x)g(x)\n24.) h(x)=\\frac{f(x)}{g(x)}\n25.) h(x)=\\frac{g(x)}{f(x)}\n28.) h(x)=\\frac{f(x)+2}{-3g(x)}

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})
  2. 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)
  3. 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)
  4. 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}})
  5. 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})
  6. 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}})