Critical Points of f(x)=\frac{\sqrt[3]{x-4}}{x-1} - Max/Min Value

[Lynne]

Homework Statement

Determine function $$f(x)=\frac{\sqrt[3]{x-4}}{x-1}$$ critical points and find max and min value in given interval $$[2; 12]$$

The Attempt at a Solution

1) I've to find derivative:

$$f(x)'=\frac{(\sqrt[3]{x-4})' (x-1)-(\sqrt[3]{x-4})(x-1)'}{(x-1)^2}=$$ $$\dfrac{\dfrac{1}{3} (x-4)^{\frac{-2}{3}} (x-4)'(x-1) - (x-4)^{\frac{1}{3}}}{(x-1)^2}=$$ $$\dfrac{\dfrac{1}{3} (x-4)^{\frac{-2}{3}}(x-1) - (x-4)^{\frac{1}{3}}}{(x-1)^2}$$

2)Critical points are:

a) 2 and 12

b) $$\neg f'(x)$$ if x=1

c) $$\dfrac{\dfrac{1}{3} (x-4)^{\frac{-2}{3}}(x-1) - (x-4)^{\frac{1}{3}}}{(x-1)^2}=0$$

here I stopped

 

[Dick]

You are missing a critical point. You have to keep simplifying that expression to see it. Try multiplying numerator and denominator by (x-4)^(2/3).