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Using Critical Points

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Chat about Calculus

What are critical points?

I will first explain critical points theoretically. One type of critical point is a local minimum or maximum in the graph. The equation y = x2 has a local minimum at (0, 0). It is a point at which the curve is lower than the surrounding points on either side. In this case, the minimum is also absolute, because it is the lowest part of the whole curve.

 

 

The second sort of critical point is an inflection point. This is a point in the graph where on one side, the slop is increasing, and on the other side, the slope is decreasing. This can happen whether the curve is increasing or decreasing. Look in the following graph at the first inflection point. Before it, the slope started from flat and increased to a slant, and after the point, the slope decreases back to flat. The inflection point is the exact point at which this transition occurs. We refer to this as a change in concavity. Before the point, it is concave up, and after, it is concave down. After the second inflection point, it is concave up again. 

You should be able to realize that if the graph is continuous and smooth, between every min and max there must be an inflection point. The converse is not necessarily true. 

I can now explain mathematically what these points are. Critical points are the areas at which y' and y" are equal to 0, simply put. 

The local maximums or minimums can be found by setting the first derivative to 0. This works because when the slope is 0, the graph is flat. If the graph is flat, it is almost always because it was going down, and now it's going up, making a minimum, or the opposite. If the object was moving upward, it is switching direction, to go downward, and for a split second the velocity is 0. This should be obvious from the graph below. At the local max and local min points, the derivative will be 0. So make it equal to 0, and see what x-values emerge. 

Inflection points are found by setting the second derivative to 0. If the first derivative measure the rate of change of y, then the second derivative measures the rate of change in y'. This measures the rate at which the slop is changing. If the second derivative is positive, the the slope is changing at a faster and faster pace. If there is a point at which the second derivative becomes 0, and then negative, the angle of the slope will stop becoming steeper, and it will then become less and less steep, possible until the curve is flat. And further. The slope can decrease to the point where it is negative, and the curve will be decreasing. When y" is positive, the graph is concave up. When it is negative, the graph is concave down. If there is a point where it is 0, then that means at that point the graph is switching from concave up to concave down, or vice versa. Look back at the graph above. 

 

Velocity and Acceleration

The function f(x) can be referring to the displacement of a particle or object. Displacement is the distance traveled from the starting point. The x-axis is time, and the y-axis is the distance moved. It can be thought of as a ball thrown directly upward, and you are plotting its position against time. Of course, that would have a particular curve. Now the derivative of this kind of function would be the velocity, because it would be plotting the rate of change against time. Remember that the derivative of a function is an equation for the slope at any point that you plug x in? Well, in an equation of displacement, the slope is the velocity, or the speed at which the displacement changes over time. So y’ is the velocity. That's what velocity is! Speed. Speed is how quickly something moves, or in other words, how quickly displacement changes. Using the same logic, you can see that y” is the acceleration, because it is the derivative of the velocity, or the rate of change of velocity over time. If the velocity is increasing then the acceleration is positive. I just explained how that sort of thing works by inflection points above.

Here’s an example:

y = x2 – 4x

I will take the derivative and second derivative:

y’ = 2x – 4

y” = 2

In this example, there is a constant acceleration of 2 for all values of x. This makes sense logically. As you can see, the graph begins with a negative velocity, (displacement is decreasing) but it begins to slow its backward movement, which is reverse deceleration, or acceleration. This constant acceleration eventually brings velocities positive.

We would like to find the local maximums and/or minimums. This would be a local minimum, and is seen in our example. I will now set y’ to 0.

y’ = 2x – 4 = 0         2x = 4         x = 2

So when x is equal to 2 the velocity is 0 and the object has reached its minimum value. How did I know it was a minimum value and not a maximum value? Because at the point x = 2 the acceleration is positive. In fact, in this entire equation the acceleration is positive; it always equals 2. When the acceleration is positive, it means the velocity is going upward, which means that it must have been negative before and is positive now. That means it is a minimum. The term for this is concave up. When a graph is concave up, it means the slant is slowly getting higher, or less negative. When the acceleration is negative, it is concave down, because that will be the shape of the graph at that point. There will be a maximum, and the slant of the graph is going to be on a downward trend. I'm sorry to repeat myself here. I don't want to insult anybody's intelligence. 

(Note that in rare cases there will be an inflection point on the same spot as the first derivative is 0, and in that case, the point is not a min or max, but the graph slows down at that point to a slant of 0, and then continues in the same direction it was going before.)

 

Drawing a graph using critical points

Firstly, what are critical points? Allow me to repeat a bit. These are all points where y’ and y” are equal to 0. When y’ is equal to 0 you have local maximums and minimums, as explained. When y” equals 0, you have inflection points, or changes in concavity. While y” is positive, it is concave up, and while it’s negative, the graph is concave down. So in between, when it is 0, the graph is switching from concave up to down, or the other way around.

The way that you figure out what to do with an equation is by using the following chart: (If you end up needing another column or two do not worry)

x

-∞

 

 

 

y

 

 

 

 

 

 

 

 

 

 

y’’

 

 

 

 

 

The top line refers to what x is. The next the rows are y, y’ and y”. You always want to know what y is at -∞ and ∞. For this you use limit of the equation as x goes to both, respectively. You also want to set y’ and y” to zero and fill in the value of x at which this occurs. Here is an equation:

y = x3 – 6x2 + 9x

y’ = 3x2 – 12 x + 9

y” = 6x – 12

I will set y’ to 0 and plug it into the graph, and set y” to 0 and plug it in.

3x2 – 12x + 9 = 0        3(x2 – 4x + 3) = 0         (x – 3)(x – 1) = 0        x = 3, 1

6x – 12 = 0         6x = 12      x = 2

Now I will plug in all values of x at which there are critical points:

x

-∞

1

2

3

y

 

 

 

 

 

 

0

 

0

 

y’’

 

 

0

 

 

Next I will Find all values of y for every critical point so I can know the full (x, y) coordinate at which these occur. I will also find -∞ and ∞.

Lim xΰ-∞ Y = x3 – 6x2 + 9x = x3 = -∞

Lim xΰ∞ Y = x3 – 6x2 + 9x = x3 = ∞

y = x3 – 6x2 + 9x = 1 – 6 + 9 = 4 (When x = 1)

y = x3 – 6x2 + 9x = 8 – 24 + 18 = 2 (When x = 2)

y = x3 – 6x2 + 9x = 27 – 54 + 27 = 0 (When x = 3)

x

-∞

1

2

3

y

-∞

4

2

0

 

0

 

0

 

y’’

 

 

0

 

 

You can already see from the chart that y is rising from negative infinity up to 4, and goes down to 0 when x is 3, and then rises to infinity. In between, it switched concavity at the point (2, 2). It is clear that  at point (1, 4) the graph is concave down, since that is a maximum, but I’ll figure it out anyway by checking y” at that point. I’ll also check concavity at (3, 0) to make sure that’s a minimum.

y” = 6x – 12 = 6 – 12 = -6 (When x is 1)

y” = 6x – 12 = 18 – 12 = 6 (When x is 3)

x

-∞

1

2

3

y

-∞

4

2

0

 

0

 

0

 

y’’

 

-6

0

6

 

Now you have every part of the graph you need to fill in to solve the graph. Here is a picture of the graph, as you should draw it. Once again, excuse the sloppiness. :)

Incidentally, once you know the (x, y) coordinates of all the critical points, and you know the direction the graph goes in to negative infinity and infinity, you can immediately figure out what the graph looks like without have to check concave up/down, etc. Plot the 3 points, and draw a line coming in from (-∞, ∞), and draw a line leaving to (∞, ∞). in the center area where you have the points, it obviously climbs to the first point, comes through the second, and loops back up at the third. There cannot be any extra squiggles in the graph, because if there were, each squiggle would have another set of max, min, and inflection points. So draw the simplest possible curve for it, and that will be correct. 

There are, incidentally, much more difficult curves possible. I will try one now.

y = 1/x - 4x2 

y' = -1/x2 - 8x

y'' = 2/x3 - 8

Set y' to 0:  -1/x2 - 8x = 0      -1/x2 - 8x3/x2 = 0      -(8x3 + 1)/x2 = 0
A fraction is 0 when the numerator is 0, so  8x3 + 1 = 0      x3 = -1/8      x = -1/2

Set y'' to 0:  2/x3 - 8 = 0      2/x3 = 8     2/8 = x3     x = 3√4

Lim x-->-∞ y = 1/x - 4x2 = 1/-∞ - 4(-∞)2 = 0 - ∞ = -∞
Lim x-->∞ y = 1/x - 4x2 = 1/∞ - 4(∞)2 = 0 - ∞ = -∞

y = 1/(-1/2) - 4(-1/2)2 = -3
y = 1/(3√4) - 4(3√4)2 = [1 - 4(3√4)2(3√4)1]/3√4 = -15/3√4

x

-∞

-1/2 

3√4

y

-∞

-3

-15/3√4

-∞

 

0

 

 

y’’

 

 

0

 

This function demonstrates the shortcomings of the chart method I use to solve these problems. Everything looks great, but something's missing. What happens when the graph is near 0? You should worry about domain in any of these problems. The domain of x refers to all the possible values it can have. In this case, x cannot be 0. It is not in the domain. If you try to put 0 in the original equation, you end up dividing by 0. We will therefore have to make special consideration for this. We must use some method to find out what happens in the 0 region. For this problem, I have two possible ways. 

The first way requires no calculation. At small numbers, the second term will be very small, and of little influence on the value of the equation. The first term will dominate. So I think to myself, what does the 1/x graph look like? The answer is:

So I know that near 0, our equation will look like this. It will go down to negative infinity, and then come down from positive infinity.

The second way involves not realizing that. Imagine putting a tiny number into 1/x - 4x2. The first term will get large, as you will be dividing by a tiny number, and the second term will near 0. The equation overall will be very big. Then put in a tiny negative number. The same thing will happen, except the first term will be very large and negative. I can assume that the smaller the number I put in, the bigger it will get. This is a vertical asymptote. (For more on this, see Limits

Whichever you try, we now know 4 lines, and two points.

This turns out to be an unusual looking graph. It looks like abstract art. Let me fill in the rest:

 

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Copyright 2004 Bruce
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