suppose that $f(4)=2$, $g(4)=3$, $f(4)= - 4$, and $g(4)=5$. find $h(4)$.\n(a) $h(x)=2f(x)+3g(x)$\n$h(4)=squar…

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: [ \begin{align*} h'(4)&=\frac{g'(4)(f(4)+g(4))-g(4)(f'(4)+g'(4))}{(f(4)+g(4))^{2}}\ &=\frac{5\times(2 + 3)-3\times((-4)+5)}{(2 + 3)^{2}}\ &=\frac{5\times5-3\times1}{25}\ &=\frac{25 - 3}{25}\ &=\frac{22}{25} \end{align*} ]

Answer:

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