Question

suppose that f(5) = 1, f(5) = 9, g(5) = -5, and g(5) = 7. find the following values.
(a) (fg)(5)
(b) (\frac{f}{g})(5)
(c) (\frac{g}{f})(5)

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) \(52\)