if ( f(x)=\frac{g(x)}{h(x)} ), then ( f^{prime}(4)= )

Explanation:

Step1: Find the equations of (g(x)) and (h(x))

For (g(x)): Using the two - point form (y - y_1=m(x - x_1)). The points ((0,1)) and ((3,7)). The slope (m=\frac{7 - 1}{3-0}=2). The equation (g(x)=2x + 1) (since (y=mx + b) and (b = 1) when (x = 0)). For (h(x)): For (x\in[3,5]), using the two - point form with points ((3,5)) and ((5,2.5)). The slope (m=\frac{2.5-5}{5 - 3}=-\frac{5}{4}). The equation (h(x)=-\frac{5}{4}x+\frac{35}{4}) (using (y - y_1=m(x - x_1)) with ((x_1,y_1)=(3,5)): (y-5=-\frac{5}{4}(x - 3)), (y=-\frac{5}{4}x+\frac{15}{4}+5=-\frac{5}{4}x+\frac{35}{4})). (g(4)=2\times4 + 1=9), (h(4)=-\frac{5}{4}\times4+\frac{35}{4}=\frac{-20 + 35}{4}=\frac{15}{4}).

Step2: Use the quotient rule

The quotient rule states that if (f(x)=\frac{u(x)}{v(x)}), then (f^\prime(x)=\frac{u^\prime(x)v(x)-u(x)v^\prime(x)}{v(x)^2}). Here (u(x)=g(x)), (u^\prime(x)=2); (v(x)=h(x)), (v^\prime(x)=-\frac{5}{4}). (f^\prime(4)=\frac{g^\prime(4)h(4)-g(4)h^\prime(4)}{h(4)^2}). Substitute (g^\prime(4) = 2), (g(4)=9), (h(4)=\frac{15}{4}), (h^\prime(4)=-\frac{5}{4}) into the formula: [ \begin{align*} f^\prime(4)&=\frac{2\times\frac{15}{4}-9\times(-\frac{5}{4})}{(\frac{15}{4})^2}\ &=\frac{\frac{30}{4}+\frac{45}{4}}{\frac{225}{16}}\ &=\frac{\frac{30 + 45}{4}}{\frac{225}{16}}\ &=\frac{\frac{75}{4}}{\frac{225}{16}}\ &=\frac{75}{4}\times\frac{16}{225}\ &=\frac{4}{3} \end{align*} ]

Answer:

(\frac{4}{3})