Question

consider the function $f(x)=8(x - 4)^{2/3}$. for this function there are two important intervals: $(-infty,a)$ and $(a,infty)$ where $a$ is a critical number. $a$ is for each of the following intervals, tell whether $f(x)$ is increasing or decreasing. $(-infty,a)$: select an answer $ (a,infty)$: select an answer for each of the following intervals, tell whether $f(x)$ is concave up or concave down. $(-infty,a)$: select an answer $ (a,infty)$: select an answer question help: video message instructor

Explanation:

Step1: Find the first derivative

Use the chain rule. If \(y = 8(u)^{2/3}\) where \(u=x - 4\), then \(\frac{dy}{du}=8\times\frac{2}{3}u^{-1/3}=\frac{16}{3}(x - 4)^{-1/3}\), and \(\frac{du}{dx}=1\). So \(f^{\prime}(x)=\frac{16}{3(x - 4)^{1/3}}\).
Set \(f^{\prime}(x)=0\), but \(\frac{16}{3(x - 4)^{1/3}} = 0\) has no solution. The derivative is undefined when \(x-4 = 0\), so \(A = 4\).

Step2: Determine increasing/decreasing

For the interval \((-\infty,4)\), pick \(x = 3\). Then \(f^{\prime}(3)=\frac{16}{3(3 - 4)^{1/3}}=\frac{16}{3(-1)}<0\), so \(f(x)\) is decreasing on \((-\infty,4)\).
For the interval \((4,\infty)\), pick \(x = 5\). Then \(f^{\prime}(5)=\frac{16}{3(5 - 4)^{1/3}}=\frac{16}{3}>0\), so \(f(x)\) is increasing on \((4,\infty)\).

Step3: Find the second derivative

\(f^{\prime}(x)=\frac{16}{3}(x - 4)^{-1/3}\), then \(f^{\prime\prime}(x)=\frac{16}{3}\times(-\frac{1}{3})(x - 4)^{-4/3}=-\frac{16}{9(x - 4)^{4/3}}\).
Since \((x - 4)^{4/3}>0\) for all \(x
eq4\), \(f^{\prime\prime}(x)<0\) for all \(x
eq4\). So \(f(x)\) is concave down on \((-\infty,4)\) and \((4,\infty)\).

Answer:

\(A = 4\)
\((-\infty,4)\): Decreasing
\((4,\infty)\): Increasing
\((-\infty,4)\): Concave down
\((4,\infty)\): Concave down