Question

suppose that $f(4)=2$, $g(4)=3$, $f(4)= - 4$, and $g(4)=5$. find $h(4)$.
(a) $h(x)=2f(x)+3g(x)$
$h(4)=square$
(b) $h(x)=f(x)g(x)$
$h(4)=square$
(c) $h(x)=\frac{f(x)}{g(x)}$
$h(4)=square$
(d) $h(x)=\frac{g(x)}{f(x)+g(x)}$
$h(4)=square$

Explanation:

Step1: Use sum - rule and constant - multiple rule for (a)

The derivative of \(h(x)=2f(x)+3g(x)\) is \(h'(x)=2f'(x)+3g'(x)\) by the sum - rule \((u + v)'=u'+v'\) and constant - multiple rule \((cf)' = cf'\). Then \(h'(4)=2f'(4)+3g'(4)\). Substitute \(f'(4)= - 4\) and \(g'(4)=5\) into the formula: \(h'(4)=2\times(-4)+3\times5=-8 + 15=7\).

Step2: Use product - rule for (b)

The product - rule states that if \(h(x)=f(x)g(x)\), then \(h'(x)=f'(x)g(x)+f(x)g'(x)\). Substitute \(x = 4\), \(f(4)=2\), \(g(4)=3\), \(f'(4)=-4\), and \(g'(4)=5\) into the formula: \(h'(4)=f'(4)g(4)+f(4)g'(4)=(-4)\times3+2\times5=-12 + 10=-2\).

Step3: Use quotient - rule for (c)

The quotient - rule states that if \(h(x)=\frac{f(x)}{g(x)}\), then \(h'(x)=\frac{f'(x)g(x)-f(x)g'(x)}{g^{2}(x)}\). Substitute \(x = 4\), \(f(4)=2\), \(g(4)=3\), \(f'(4)=-4\), and \(g'(4)=5\) into the formula: \(h'(4)=\frac{f'(4)g(4)-f(4)g'(4)}{g^{2}(4)}=\frac{(-4)\times3-2\times5}{3^{2}}=\frac{-12 - 10}{9}=-\frac{22}{9}\).

Step4: Use quotient - rule for (d)

If \(h(x)=\frac{g(x)}{f(x)+g(x)}\), by the quotient - rule \(h'(x)=\frac{g'(x)(f(x)+g(x))-g(x)(f'(x)+g'(x))}{(f(x)+g(x))^{2}}\). Substitute \(x = 4\), \(f(4)=2\), \(g(4)=3\), \(f'(4)=-4\), and \(g'(4)=5\) into the formula:

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Answer:

(a) \(7\)
(b) \(-2\)
(c) \(-\frac{22}{9}\)
(d) \(\frac{22}{25}\)