



A special
case: 


F(X) = X^{3}






The first
derivative of F(X) = X
^{3} is: 





F'(X) = 3X^{2}






To find
where the derivative is zero, set F'(X) = 0 





F'(X) = 0 = 3X^{2}



X = 0






The
second derivative of F(X) = X
^{3} is: 





F''(X)
= 6X






Now lets
see what the second derivative is doing 


where the slope of the
original equation is zero. 


That is,
let's see if it is concave up (a smile) or concave down (a frown): 


The slope
was zero when X = 0 so: 





F''(X)
= 6X



F''(0)
= 6(0)



F''(0)
= 0






OK, so
this is why this one is a special case. 


At the
point where the slope is zero, the graph is not curving up or down. 


It has a
point of inflection. 





Let's do
a little more testing. 


Pick a
point on either side of the point of inflection 


and see what the
slope is doing at those points... 





F'(X)
= 
3X^{2} 
F'(X)
= 
3X^{2} 
F'(X)
= 
3(1)^{2} 
F'(X)
= 
3(1)^{2} 
F'(X)
= 
3 
F'(X)
= 
3 
the
slope is positive (up) 
the
slope is positive (up) 






So the
slope is positive for all X values except when X = 0 


That
would make the graph look something like this: 





Let's
check and see what the second derivative is doing at X = 1 and X =
1 ... 





F''(X)
= 
6X 
F''(X)
= 
6X 
F''(X)
= 
6(1) 
F''(X)
= 
6(1) 
F''(X)
= 
 6 
F''(X)
= 
6 
concave
down 
concave
up 






Which
means the graph looks like this ... 





The frown
part gets just to where the slope is zero, 


and then it turns into
the smile part. 





A point
here is that just because there is a point where the slope is
zero, 


doesn't
mean that there HAS to be a point that is a relative minimum or
maximum. 





Another
special case happens when the denominator of an equation 


is equal to
zero. 


This
happens a lot if there is a variable in the denominator. 





If we can
factor out the terms that make the denominator zero, 


the graph will
have a hole. 


In the
equation: 








The
denominator will equal zero when X = 0. 





We can
factor an X out of the numerator and the denominator, 


so that
this equation becomes: 





F(X)
= X^{2}






This is
just a parabola. 


But the
original equation has a problem when X = 0, 


so the
graph of the equation is a parabola with a hole at X = 0: 











The other
side of the story is when the zero in the denominator 


can't be
factored out. 


When that
happens, 


there
will be a vertical asymptote at the variable value 


that makes the
denominator equal zero. 


In the
equation: 


F(X)
= 
1 
also
known as F(X) = X^{1} 

X 






The X in
the denominator can't be factored out. 


That
means there will be a vertical asymptote when X = 0 





Let's do
some calculus on this equation: 


The first
and second derivatives of this equation are: 





F'(x)
= 
1 
also
known as F'(X) = X^{2} 

X^{2} 



F''(x)
= 
2 
also
known as F''(X) = 2X^{3} 

X^{3} 






There is
no place where any of these functions (F(X),
F'(X).
F''(X)) will
equal zero. 


Since
there are only numbers, not variables, on top. 


But all
of them are undefined at X = 0. 





We can
just pick any point to the left of zero, and any point to the right
of zero. 


Then
check for slope and concavity at these points. 


All
points on that side of the X = 0 point will have the same slope and
concavity. 





Here we
go. Always use the easiest points you can. 


Let's use
1 and +1 


First 1
... 


F'(X)
= 
1 
F''(X)
= 
2 


X^{2} 
X^{3} 

F'(1)
= 
1 
F''(1)
= 
2 


(1)^{2} 
(1)^{3} 




F'(1)
= 
1 
F''(1)
= 
2 


1 
1 




F'(1)
= 
1 
F''(1)
= 
2 




the
slope is negative (down) 
concave
down (frown) 






Now +1 ...



F'(X)
= 
1 
F''(X)
= 
2 


X^{2} 
X^{3} 




F'(1)
= 
1 
F''(1)
= 
2 


(1)^{2} 
(1)^{3} 




F'(1)
= 
1 
F''(1)
= 
2 


1 
1 




F'(1)
= 
1 
F''(1)
= 
2 




the
slope is negative (down) 
concave
up (smile) 






So the
graph looks like this: 








copyright 2005 Bruce Kirkpatrick 
