let f(x)=-1/5(x - 8)^(5/3)-2(x - 8)^(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 - 8), then (f(x)=-\frac{1}{5}u^{\frac{5}{3}}-2u^{\frac{2}{3}}-1). Using the chain - rule (\frac{df}{dx}=\frac{df}{du}\cdot\frac{du}{dx}), and (\frac{du}{dx} = 1). (f^\prime(x)=-\frac{1}{5}\times\frac{5}{3}(x - 8)^{\frac{2}{3}}-2\times\frac{2}{3}(x - 8)^{-\frac{1}{3}}) (f^\prime(x)=-\frac{1}{3}(x - 8)^{\frac{2}{3}}-\frac{4}{3}(x - 8)^{-\frac{1}{3}}=-\frac{1}{3}(x - 8)^{-\frac{1}{3}}[(x - 8)+4]=-\frac{1}{3}(x - 8)^{-\frac{1}{3}}(x - 4))

Step2: Find the critical points

Set (f^\prime(x)=0), then (-\frac{1}{3}(x - 8)^{-\frac{1}{3}}(x - 4)=0). The critical points occur when (x - 4 = 0) (since ((x - 8)^{-\frac{1}{3}}\neq0) for any real (x\neq8)), so (x = 4), and (x = 8) is a point where the derivative is undefined.

Step3: Test the intervals

Consider the intervals ((-\infty,4)), ((4,8)) and ((8,\infty)).

• For (x\in(-\infty,4)), let (x = 3). Then (f^\prime(3)=-\frac{1}{3}(3 - 8)^{-\frac{1}{3}}(3 - 4)=-\frac{1}{3}(-5)^{-\frac{1}{3}}(-1)<0).
• For (x\in(4,8)), let (x = 5). Then (f^\prime(5)=-\frac{1}{3}(5 - 8)^{-\frac{1}{3}}(5 - 4)=-\frac{1}{3}(-3)^{-\frac{1}{3}}(1)>0).
• For (x\in(8,\infty)), let (x = 9). Then (f^\prime(9)=-\frac{1}{3}(9 - 8)^{-\frac{1}{3}}(9 - 4)=-\frac{5}{3}<0).

The function (f(x)) is increasing on the interval ((4,8)).

Answer:

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