let f(x)=−1/5(x - 4)^(5/3)−2(x - 4)^(2/3)−1. find the interval (a,b) where f increases. as your answer…

Explanation:

Step1: Find the derivative of (f(x))

Let (u = x - 4), then (f(u)=-\frac{1}{5}u^{\frac{5}{3}}-2u^{\frac{2}{3}}-1). Using the power - rule ((x^n)^\prime=nx^{n - 1}), (f^\prime(u)=-\frac{1}{5}\times\frac{5}{3}u^{\frac{2}{3}}-2\times\frac{2}{3}u^{-\frac{1}{3}}=-\frac{1}{3}u^{\frac{2}{3}}-\frac{4}{3u^{\frac{1}{3}}}=\frac{-u - 4}{3u^{\frac{1}{3}}}). Substituting back (u = x - 4), we get (f^\prime(x)=\frac{-(x - 4)-4}{3(x - 4)^{\frac{1}{3}}}=\frac{-x}{3(x - 4)^{\frac{1}{3}}}).

Step2: Find the critical points

Set (f^\prime(x)=0), then (-x = 0), so (x = 0). Also, (f^\prime(x)) is undefined when (x = 4).

Step3: Determine the sign of (f^\prime(x)) in intervals

Consider the intervals ((-\infty,0)), ((0,4)) and ((4,\infty)). For (x\in(-\infty,0)), let (x=-1), then (f^\prime(-1)=\frac{-(-1)}{3(-1 - 4)^{\frac{1}{3}}}=\frac{1}{3(-5)^{\frac{1}{3}}}<0). For (x\in(0,4)), let (x = 1), then (f^\prime(1)=\frac{-1}{3(1 - 4)^{\frac{1}{3}}}=\frac{-1}{3(-3)^{\frac{1}{3}}}>0). For (x\in(4,\infty)), let (x = 5), then (f^\prime(5)=\frac{-5}{3(5 - 4)^{\frac{1}{3}}}=-\frac{5}{3}<0). The function (f(x)) is increasing on the interval ((0,4)).

Answer:

(0 + 4=4)