let f(x)=(x - 2)x^{2/3}. find the interval (a,b) where f decreases. as your answer, please input a + b.

Explanation:

Step1: Find the derivative

Use the product - rule $(uv)^\prime = u^\prime v+uv^\prime$, where $u = x - 2$ and $v=x^{2/3}$. $u^\prime=1$ and $v^\prime=\frac{2}{3}x^{-1/3}$. So $f^\prime(x)=1\times x^{2/3}+(x - 2)\times\frac{2}{3}x^{-1/3}=\frac{3x+(2x - 4)}{3x^{1/3}}=\frac{5x - 4}{3x^{1/3}}$.

Step2: Find the critical points

Set $f^\prime(x)=0$, then $\frac{5x - 4}{3x^{1/3}} = 0$. The numerator gives $5x-4 = 0$, so $x=\frac{4}{5}$. Also, $f^\prime(x)$ is undefined at $x = 0$.

Step3: Determine the sign of the derivative

Test the intervals $(-\infty,0)$, $(0,\frac{4}{5})$, and $(\frac{4}{5},\infty)$. For $x\in(-\infty,0)$, let $x=-1$, then $f^\prime(-1)=\frac{-5 - 4}{-3}=3>0$. For $x\in(0,\frac{4}{5})$, let $x=\frac{1}{2}$, then $f^\prime(\frac{1}{2})=\frac{\frac{5}{2}-4}{\frac{3}{2^{1/3}}}=\frac{-\frac{3}{2}}{\frac{3}{2^{1/3}}}<0$. For $x\in(\frac{4}{5},\infty)$, let $x = 1$, then $f^\prime(1)=\frac{5 - 4}{3}=\frac{1}{3}>0$. So $f(x)$ is decreasing on the interval $(0,\frac{4}{5})$.

Answer:

$\frac{4}{5}$