Critical Points

Question: Why do we have to look at places where f is not differentiable to find local maxes and mins?  Why don't the points where f ' = 0 suffice?

Answer: Look at the function f(x) = x^(2/3)(x^3 - 1)

In[6]:=

D[x^(2/3) (x^3 - 1), x]

Out[6]=

3 x^(8/3) + (2 (-1 + x^3))/(3 x^(1/3))

In[7]:=

Simplify[%]

Out[7]=

(-2 + 11 x^3)/(3 x^(1/3))

So we see the critical points are x = 2/11^(1/3) (where the numerator is zero) and x = 0 (where the denominator is zero).

In[9]:=

Plot[x^2^(1/3) (x^3 - 1), {x, -3, 3}, PlotStyleRGBColor[0, 0, 1]]

[Graphics:HTMLFiles/index_11.gif]

Out[9]=

⁃Graphics⁃

Notice that the critical point 0 turns out to be a local maximum, and the other one is a local minimum.

Created by Mathematica  (April 6, 2005)