The Graph of f

Date: Wed, 3 Jul 1996 18:22:11 -0400 (EDT) 
From: Anonymous
Subject: function f(x)=3x^(1/3)-x

Consider the function f(x)=3x^(1/3)-x. 

Where are its local and absolute maxima and minima? 
What are the x-intercepts of the graph of f? 
Does its graph have inflection points? 
Are there asymptotes? 
Where is the graph concave upward or downward? 
Where is the function increasing/ decreasing? 
Does the graph exhibit any symmetries? 
Sketch the graph.

Thanking you ahead of time, Dr. Math.

Date: Thu, 4 Jul 1996 12:43:49 -0400 (EDT) 
From: Dr. Anthony
Subject: Re: function f(x)=3x^(1/3)-x

We note that this is an odd function of x, and so will have point 
symmetry about the origin.

f'(x) = 3(1/3)x^(-2/3) - 1 = 1/x^(2/3) - 1 

So f'(x) = 0 when x^(2/3) = 1, that is x = 1 or x = -1 

The f(1) = 3-1 = 2 and f(-1) = -3 -(-1) = -2 

So turning points at (1,2) (max.point) and (-1,-2) (min.point) 

When x=0 then y=0, so curve passes through the origin, and f'(x) is 
infinite, so curve is vertical at the origin 

When y=0 we also have 3x^(1/3) = x so 
3 = x^(2/3) or 
x = 3^(3/2) x = +or- 5.196 
The curve cuts the x axis at +and- 5.196 

When x -> infinity f'(x) -> 1/infin - 1 -> -1, so the curve tends to a 
slope -1 as x increases.

The curve has an S shape between (-1,-2), (0,0) and (1,2), the first 
and last of these being turning points, and the origin (0,0), a point 
where the curve is vertical. From (1,2) the curve slopes down as 
x increases, crosses the x axis at 5.196 and tends to a slope -1 as x 
tends to infinity . From (-1,-2) the curve bends up as x decreases, 
crosses the x axis at -5.196 and tends to a slope -1 as x tends to 
-infinity. 

There is one point of inflection (f''(x)=0) and that is at the origin. 
There are no asymptotes.

-Doctor Anthony, The Math Forum