Question

homework3: problem 26
(1 point)
given that:
$f(-5)=4,f(-5)=4,$
$g(-5)=1,g(-5)=3,$
calculate the following:
$(fg)(-5)=\square$
$(f/g)(-5)=\square$
$(g/f)(-5)=\square$
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Explanation:

Step1: Use the product rule

The product rule states that \((fg)^\prime(x)=f^\prime(x)g(x)+f(x)g^\prime(x)\).
Substitute \(x = - 5\):
\((fg)^\prime(-5)=f^\prime(-5)g(-5)+f(-5)g^\prime(-5)\)
\(=4\times1 + 4\times3\)
\(=4 + 12\)
\(=16\)

Step2: Use the quotient rule

The quotient rule states that \((\frac{f}{g})^\prime(x)=\frac{f^\prime(x)g(x)-f(x)g^\prime(x)}{g(x)^2}\).
Substitute \(x=-5\):
\((\frac{f}{g})^\prime(-5)=\frac{f^\prime(-5)g(-5)-f(-5)g^\prime(-5)}{g(-5)^2}\)
\(=\frac{4\times1-4\times3}{1^2}\)
\(=\frac{4 - 12}{1}\)
\(=-8\)

Step3: Use the quotient rule again

The quotient rule states that \((\frac{g}{f})^\prime(x)=\frac{g^\prime(x)f(x)-g(x)f^\prime(x)}{f(x)^2}\).
Substitute \(x = - 5\):
\((\frac{g}{f})^\prime(-5)=\frac{g^\prime(-5)f(-5)-g(-5)f^\prime(-5)}{f(-5)^2}\)
\(=\frac{3\times4-1\times4}{4^2}\)
\(=\frac{12 - 4}{16}\)
\(=\frac{8}{16}\)
\(=\frac{1}{2}\)

Answer:

\((fg)^\prime(-5)=16\)
\((f/g)^\prime(-5)=-8\)
\((g/f)^\prime(-5)=\frac{1}{2}\)