Question

1. 0 / 1 points let $h(x) = \frac{f(x)}{1 + g(x)}$. given $f(5) = 1$, $f(5) = -2$, $g(5) = 3$, and $g(5) = -4$, evaluate $h(5)$. $h(5) = $

Explanation:

Step1: Apply Quotient Rule

The quotient rule states that if \( h(x)=\frac{u(x)}{v(x)} \), then \( h'(x)=\frac{u'(x)v(x)-u(x)v'(x)}{[v(x)]^2} \). Here, \( u(x) = f(x) \) and \( v(x)=1 + g(x) \). So, \( u'(x)=f'(x) \) and \( v'(x)=g'(x) \). Thus, \( h'(x)=\frac{f'(x)(1 + g(x))-f(x)g'(x)}{(1 + g(x))^2} \).

Step2: Substitute \( x = 5 \)

Substitute \( x = 5 \) into the derivative formula. We know \( f(5)=1 \), \( f'(5)=-2 \), \( g(5)=3 \), and \( g'(5)=-4 \). So, \( 1 + g(5)=1 + 3 = 4 \), \( f'(5)(1 + g(5))=-2\times4=-8 \), \( f(5)g'(5)=1\times(-4)=-4 \). Then the numerator is \( -8-(-4)=-8 + 4=-4 \), and the denominator is \( (1 + g(5))^2=4^2 = 16 \).

Step3: Calculate \( h'(5) \)

Now, \( h'(5)=\frac{-4}{16}=-\frac{1}{4} \).

Answer:

\( -\frac{1}{4} \)