Question

question 7 estimate the instantaneous rate of change of $g(t) = \frac{4}{t + 4}$ at the point $t = -2$ your answer should be accurate to at least 3 decimal places. question help: message instructor submit question jump to answer

Explanation:

Step1: Recall the Quotient Rule

The quotient rule for differentiation states that if \( g(t) = \frac{u(t)}{v(t)} \), then \( g'(t) = \frac{u'(t)v(t) - u(t)v'(t)}{[v(t)]^2} \). For \( g(t) = \frac{4}{t + 4} \), we have \( u(t) = 4 \) (so \( u'(t) = 0 \)) and \( v(t) = t + 4 \) (so \( v'(t) = 1 \)).

Step2: Apply the Quotient Rule

Substitute into the quotient rule: \( g'(t) = \frac{0 \cdot (t + 4) - 4 \cdot 1}{(t + 4)^2} = \frac{-4}{(t + 4)^2} \).

Step3: Evaluate at \( t = -2 \)

Substitute \( t = -2 \) into \( g'(t) \): \( g'(-2) = \frac{-4}{(-2 + 4)^2} = \frac{-4}{(2)^2} = \frac{-4}{4} = -1 \). Wait, that's exact. But let's check again. Wait, \( v(t)=t + 4 \), so \( v(-2)=-2 + 4 = 2 \), \( [v(-2)]^2 = 4 \), \( u'(t)v(t)-u(t)v'(t)=0(t + 4)-41=-4 \), so \( g'(-2)=\frac{-4}{4}=-1.000 \) (to three decimal places).

Answer:

\( -1.000 \)