Question

graph the function ( f(x) = 16(x - 16)^{\frac{2}{3}} ), then find the intervals of increase and decrease.
for the following questions, state all your answers in interval notation. reduce all fractions and radicals. enter exact numbers. if the interval does not exist, enter dne.
a.) find the intervals of increase.
b.) find the intervals of decrease.

Explanation:

Step1: Find the derivative of the function

Use the chain rule. If \(y = 16u^{\frac{2}{3}}\) and \(u=x - 16\), then \(\frac{dy}{du}=16\times\frac{2}{3}u^{-\frac{1}{3}}=\frac{32}{3}(x - 16)^{-\frac{1}{3}}\), and \(\frac{du}{dx}=1\). By the chain rule \(f^\prime(x)=\frac{32}{3}(x - 16)^{-\frac{1}{3}}=\frac{32}{3\sqrt[3]{x - 16}}\).

Step2: Find the critical point

Set the denominator of \(f^\prime(x)\) equal to zero. \(3\sqrt[3]{x - 16}=0\) gives \(x = 16\). The domain of \(f(x)\) is all real numbers (\(x\in(-\infty,\infty)\)) and the derivative \(f^\prime(x)\) is undefined at \(x = 16\).

Step3: Test intervals

• For the interval \((-\infty,16)\), let \(x=15\). Then \(f^\prime(15)=\frac{32}{3\sqrt[3]{15 - 16}}=\frac{32}{3(- 1)}=-\frac{32}{3}<0\).
• For the interval \((16,\infty)\), let \(x = 17\). Then \(f^\prime(17)=\frac{32}{3\sqrt[3]{17 - 16}}=\frac{32}{3(1)}=\frac{32}{3}>0\).

Answer:

a. \((16,\infty)\)
b. \((-\infty,16)\)