suppose that f(5) = 1, f(5) = 9, g(5) = -5, and g(5) = 7. find the following values.\n(a) (fg)(5)\n(b)…

Explanation:

Step1: Recall product - rule

The product - rule states that ((fg)'(x)=f'(x)g(x)+f(x)g'(x)). Substitute (x = 5): ((fg)'(5)=f'(5)g(5)+f(5)g'(5)). Given (f(5)=1), (f'(5)=9), (g(5)= - 5), and (g'(5)=7). ((fg)'(5)=9\times(-5)+1\times7=-45 + 7=-38).

Step2: Recall quotient - rule

The quotient - rule states that ((\frac{f}{g})'(x)=\frac{f'(x)g(x)-f(x)g'(x)}{g^{2}(x)}). Substitute (x = 5): ((\frac{f}{g})'(5)=\frac{f'(5)g(5)-f(5)g'(5)}{g^{2}(5)}). (g^{2}(5)=(-5)^{2}=25), (f'(5)g(5)-f(5)g'(5)=9\times(-5)-1\times7=-45 - 7=-52). So ((\frac{f}{g})'(5)=\frac{-52}{25}).

Step3: Recall quotient - rule for (\frac{g}{f})

The quotient - rule for ((\frac{g}{f})'(x)=\frac{g'(x)f(x)-g(x)f'(x)}{f^{2}(x)}). Substitute (x = 5): ((\frac{g}{f})'(5)=\frac{g'(5)f(5)-g(5)f'(5)}{f^{2}(5)}). (f^{2}(5)=1^{2}=1), (g'(5)f(5)-g(5)f'(5)=7\times1-(-5)\times9=7 + 45=52). So ((\frac{g}{f})'(5)=52).

Answer:

(a) (-38) (b) (-\frac{52}{25}) (c) (52)