Critical points

1. May 9, 2009

Lynne

1. The problem statement, all variables and given/known data
Determine function $$f(x)=\frac{\sqrt[3]{x-4}}{x-1}$$ critical points and find max and min value in given interval $$[2; 12]$$

3. 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

2. May 9, 2009

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).