Question

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:

$$
LATEXBLOCK0
$$

Answer:

\(\frac{4}{3}\)