| 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. | ||
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| 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. | ||
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| 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. | ||
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| 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 | ||
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| OK. let's do one for real. | ||
| Example: | ||
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F(X) = X2 |
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| This equation has a derivative, | ||
| we'll call it a first derivative, of: | ||
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F'(X) = 2X |
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| So the second derivative is: | ||
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F''(X) = 2X |
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| Since 2 is positive no matter what the value of X is, | ||
| the graph of the original equation will be a smile. | ||
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| 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: | ||
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F(X) = X3 - 3X2 - 24X + 5 |
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| How do you find the shape of this graph? | ||
| First, find the first derivative... | ||
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F'(X) = 3X2 - 6X - 24 |
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| 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: | ||
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0 = 3X2 - 6X - 24 |
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| 0 = 3(X2 - 2X - 8) | ||
| 0 = 3(X - 4)(X + 2) | ||
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So the solutions are X = 4 and X = -2. |
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At each of these points, the graph of the original equation is horizontal. |
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| 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 | ||
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F'(X) = 3X2 - 6X - 24 |
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F'(0) = 3(0)2 - 6(0) - 24 |
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F'(0) = - 24 |
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| The slope is negative, so the graph goes down in the middle. | ||
| Pick any point to the left of -2. Let's try, -3... | ||
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F'(X) = 3X2 - 6X - 24 |
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F'(-3) = 3(-3)2 - 6(-3) - 24 |
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F'(-3) = 27 + 18 - 24 |
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F'(-3) = 21 |
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| 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... | ||
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F'(X) = 3X2 - 6X - 24 |
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F'(5) = 3(5)2 - 6(5) - 24 |
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F'(5) = 75 - 30 - 24 |
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F'(5) = 21 |
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| The slope is positive so the graph goes up on the right. | ||
| So putting this all together, we get: | ||
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| 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... | ||
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F'(X) = 3X2 - 6X - 24 |
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| So | ||
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F''(X) = 6X - 6 |
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| Now, check the second derivative value at the places | ||
| where the first derivative was zero. | ||
| That is, -2 and 4 | ||
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F''(X) = 6X - 6 |
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F''(-2) = 6(-2) - 6 |
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F''(-2) = -18 |
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| So the graph is concave downward around X = -2 | ||
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F''(X) = 6X - 6 |
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F''(4) = 6(4) - 6 |
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F''(4) = 18 |
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| So the graph is concave upward around X = 4 | ||
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| 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 ... | ||
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F(X) = X3 - 3X2 - 24X + 5 |
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F(-2) = (-2)3 - 3(-2)2 - 24(-2) + 5 |
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F(-2) = -8 -12 + 48 + 5 |
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F(-2) = 33 |
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| So the relative maximum is at (-2,33) | ||
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F(X) = X3 - 3X2 - 24X + 5 |
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| F(4) = (4)3 - 3(4)2 - 24(4) + 5 | ||
| F(4) = 64 - 48 -96 + 5 | ||
| F(4) = -75 | ||
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So the relative minimum point is at (4,75) |
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| 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... | ||
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F''(X) = 6X - 6 |
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| Solve this for X | ||
| where the function value is zero... | ||
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0 = 6X - 6 |
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0 = 6(X - 1) |
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X = 1 |
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| Plugging X = 1 back into the original equation... | ||
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F(X) = X3 - 3X2 - 24X + 5 |
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| F(1) = (1)3 - 3(1)2 - 24(1) + 5 | ||
| F(1) = 1 - 3 - 24 + 5 | ||
| F(1) = -21 | ||
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So the point of inflection is (1,-21) |
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| 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 |
