Question

find the average rate of change of k(x) = -19\sqrt{x}+19 over the interval 10, 12. write your answer as an integer, fraction, or decimal rounded to the nearest tenth. simplify any fractions.

Explanation:

Step1: Recall average rate - of - change formula

$\text{Average rate of change}=\frac{k(b)-k(a)}{b - a}$, where $a = 10$, $b = 12$.

Step2: Calculate $k(10)$ and $k(12)$

$k(10)=-19\sqrt{10}+19$, $k(12)=-19\sqrt{12}+19$.

Step3: Substitute into formula

$\frac{(-19\sqrt{12}+19)-(-19\sqrt{10}+19)}{12 - 10}=\frac{-19\sqrt{12}+19 + 19\sqrt{10}-19}{2}=\frac{-19(\sqrt{12}-\sqrt{10})}{2}\approx - 3.4$.

Answer:

$-3.4$