



If you
take the derivative of a derivative 


you get something called a
second derivative (how original). 





On a
previous page, 


we said that the second derivative was the change
in velocity or speed. 


We also
said that a change in speed is acceleration. 


(And
deceleration is just acceleration with a minus sign) 





Anyway,
the second derivative also helps with graphing. 





Wherever
the second derivative of an original equation is positive, 


the graph
of that original equation curves up. 





So it
will look kind of like a smile wherever the second derivative is
positive. 


We call
this "Concave Upwards." 


See that
in some places in the smile the graph is going down 


and in some
places it is going up. 


And at
one place the slope has to be zero. 





For
values of the variable that makes the second derivative negative, 


the graph
of the original equation curves down. 





So the
graph of the original equation looks like a frown 


for whatever
values of the variable make the
second derivative negative. 


We call
this "Concave Downwards" 





Variable
values that make the second derivative equal to zero happen 


where the
graph changes from curving one way to curving the other way. 





These
points are called "Points of Inflection." 


It's no
use leaving out the "l" in the name. 


That was
funny about 50 or 60 years ago, but not now. 





OK,
here's something to think about... 


What if
the graph of the original equation is a straight line? 





In that
case, the graph is not curving up anywhere 


so the second derivative
is never positive. 


And the
graph is not curving down anywhere 


so the second derivative is not
negative. 





Any rules
we could build from this? 





How
about: Equations of straight lines all have second derivatives equal
to zero. 


That is,
for any straight line: F''(X) = 0 








OK. let's
do one for real. 





Example: 





F(X) = X^{2}



This
equation has a derivative, 


we'll
call it a first derivative, of: 





F'(X)
= 2X






So the
second derivative is: 





F''(X)
= 2X






Since 2
is positive no matter what the value of X is, 


the graph of the
original equation will be a smile. 








We could
plug in a couple of values for X and tack this smile to a set of
axis, 


but basically this is the guts of the answer. 





OK, Let's
try a real hard one: 





Example: 





F(X) = X^{3}
 3X^{2}  24X + 5






How do
you find the shape of this graph? 


First,
find the first derivative... 





F'(X)
= 3X^{2}  6X  24






The graph
of the original function will be horizontal 


where the first
derivative is zero. 


We can
find those points by setting the derivative equal to zero 


and
solving for X. 


We can
solve this for X by factoring (if we're lucky) 


or using the
quadratic formula if we're not. 


Let's try
factoring: 


0
= 3X^{2}  6X  24



0 =
3(X^{2}  2X  8) 


0 =
3(X  4)(X + 2) 





So the
solutions are X = 4 and X = 2. 


At each of
these points, the graph of the original equation is horizontal. 








In the
space BETWEEN these two points, 


the graph
will either be going up for the whole distance 


or going
down for the whole distance. 


In the
space from 4 to positive infinity, 


the graph will either be always
going up or always going down. 





The same
is true from 2 to negative infinity on the left. 





So how do
we know if it is going up or down in the three areas? 





There are
two ways we can do this: 


The first
way is to pick any point in each of the spaces 


and see
if the first derivative is positive or negative for that point. 





The easy
place to check is where X = 0 





F'(X)
= 3X^{2}  6X  24



F'(0)
= 3(0)^{2}  6(0)  24



F'(0)
=  24






The slope
is negative, so the graph goes down in the middle. 





Pick any
point to the left of 2. Let's try, 3... 





F'(X)
= 3X^{2}  6X  24



F'(3)
= 3(3)^{2}  6(3)  24



F'(3)
= 27 + 18  24



F'(3)
= 21






The slope
is positive so the graph goes up on the left. 





Finally,
we pick a point to the right of 4. Let's try 5... 





F'(X)
= 3X^{2}  6X  24



F'(5)
= 3(5)^{2}  6(5)  24



F'(5)
= 75  30  24



F'(5)
= 21






The slope
is positive so the graph goes up on the right. 





So
putting this all together, we get: 





We could
plot a few points, and pin this down pretty easily. 


But lets
look at the other way to graph this first... 





The
second way is to figure out what the graph is doing 


using the second
derivative. 


To do
that, first we need to figure out what the second derivative is... 





F'(X)
= 3X^{2}  6X  24



So 


F''(X)
= 6X  6






Now,
check the second derivative value at the places 


where the first
derivative was zero. 


That is,
2 and 4 


F''(X)
= 6X  6



F''(2)
= 6(2)  6



F''(2)
= 18






So the
graph is concave downward around X
= 2 





F''(X)
= 6X  6



F''(4)
= 6(4)  6



F''(4)
= 18






So the
graph is concave upward around X
= 4 








That
looks a lot like what we got using the first derivative stuff. 


Hey, it
should be. It's the same equation! 





There are
three special places that we would like ways to find on this graph. 





One is
the top of the "hill" on the left. This is called a
relative maximum. 


It is a
relative maximum because it is higher than any other point nearby. 


There may
be a function value on the right that is even higher than
this, 


so this
point is not necessarily an absolute maximum. 





The
second special point is the bottom of the valley on the right. 


You can
probably guess that this point is called a relative minimum. 





The last
special point is where the graph stops being concave down 


and starts
being concave up. 


This is
the point of inflection. 





Relative
minimum and relative maximum points happen 


where the slope of the
equation is zero. 





We
already found those points. They are at X = 2 and X = 4 


We just
need to find the function values at those points: 


Plugging
those values into the original equation ... 





F(X) = X^{3}
 3X^{2}  24X + 5



F(2) =
(2)^{3}
 3(2)^{2}  24(2) + 5



F(2) =
8 12 + 48 + 5



F(2) =
33






So the
relative maximum is at (2,33) 





F(X) = X^{3}
 3X^{2}  24X + 5



F(4) =
(4)^{3}
 3(4)^{2}  24(4) + 5 


F(4) =
64  48 96 + 5 


F(4) =
75 





So the
relative minimum point is at (4,75) 





Since
points of inflection happen where the second derivative is
zero, 


we just
set the second derivative equal to zero and solve for X. 


Then
we'll find the function value for that point. 


From
above, we know that... 





F''(X)
= 6X  6






Solve
this for X 


where the
function value is zero... 





0
= 6X  6



0
= 6(X  1)



X
= 1






Plugging
X = 1 back into the original equation... 





F(X) =
X^{3}
 3X^{2}  24X + 5



F(1) =
(1)^{3}
 3(1)^{2}  24(1) + 5 


F(1) =
1
 3  24 + 5 


F(1) =
21 





So the
point of inflection is (1,21)






Let's
review: 


1) The
slope can only change direction (like up to down or down to
up), 


at a point where the first derivative is zero or undefined. 





2) If
there is a point on a graph where the first derivative is zero 


and
the second derivative is positive,
the graph will have a relative maximum. 





3) If
there is a point on a graph where the first derivative is zero 


and
the second derivative is negative,
the graph will have a relative minimum. 





copyright 2005 Bruce Kirkpatrick 
