Calculus Graphing with the Second Derivative
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Finding Concavity
Graphing with the Second Derivative

 
 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) = X2

 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) = X3 - 3X2 - 24X + 5

 
 How do you find the shape of this graph?
 First, find the first derivative...
 

 F'(X) = 3X2 - 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 = 3X2 - 6X - 24

0 = 3(X2 - 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) = 3X2 - 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) = 3X2 - 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) = 3X2 - 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) = 3X2 - 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) = X3 - 3X2 - 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) = X3 - 3X2 - 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) = X3 - 3X2 - 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

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