Question

find the average rate of change of ( k(x)=\frac{-16}{x + 12} ) over the interval ( -11,-10 ).
write your answer as an integer, fraction, or decimal rounded to the nearest tenth.
simplify any fractions.

Explanation:

Step1: Recall the formula for average rate of change

The formula for the average rate of change of a function \(y = k(x)\) over the interval \([a,b]\) is \(\frac{k(b)-k(a)}{b - a}\). Here, \(a=-11\), \(b = - 10\).

Step2: Find \(k(-11)\) and \(k(-10)\)

For \(k(x)=\frac{-16}{x + 12}\), when \(x=-11\), \(k(-11)=\frac{-16}{-11 + 12}=\frac{-16}{1}=-16\).
When \(x=-10\), \(k(-10)=\frac{-16}{-10 + 12}=\frac{-16}{2}=-8\).

Step3: Calculate the average rate of change

Substitute into the formula: \(\frac{k(-10)-k(-11)}{-10-(-11)}=\frac{-8-(-16)}{-10 + 11}\).
Simplify the numerator: \(-8 + 16=8\), and the denominator: \(-10 + 11 = 1\).

Answer:

\(8\)