At the end of this session you will be able to
For the last four weeks, we learned how to differentiate product functions, quotient functions, power functions, and composite functions and learned how to differentiate when we have functions given by parametric equations and how to perform implicit differentiation. We are going to look at applications of maximum and minimum values and look at theoretical considerations of the derivative.
ABSOLUTE EXTREME VALUES
M = f (a) is the absolute maximum value for f(x) if f(a) > or = f(x) for all x-values in the domain of the function f.
m = f (a) is the absolute minimum value for f(x) if f(a) < or = f(x) for all x-values in the domain of the function f.
Frequently, mathematicians leave out the word "absolute" when discussing absolute extrema and simply use the words "maximum" or "minimum". These absolute extrema can occur either at endpoints or within an interval. Therefore, we look at absolute extrema at endpoints or at interior points. In an open interval with endpoints missing, we therefore might not have absolute extrema as illustrated in the table on page 245.
SUFFICIENT CONDITION FOR ABSOLUTE EXTREME VALUES
However, if we have a closed interval and continuity, then we always know there will always be an absolute maximum value and an absolute minimum value. That conclusion is expressed in Theorem 1 "The Extreme-Value Theorem for Continuous Functions" on page 246 of your textbook, which is sometimes called the sufficient condition for absolute extrema. What do we mean by sufficient condition? Sufficient condition tells you that hypothesis yields the conclusion, but the converse is not necessarily true. In other words, you can have extrema without continuity or without a closed interval.
Now you can go to the Session 12 assignments to continue your assignment.
At the end of this session you will be able to
LOCAL (RELATIVE) EXTREME VALUES
If x = c is not at an endpoint, then f (c) is a local (relative) maximum value for f(x) if f(c) > or = f(x) for all x-values in a small open interval around x = c.
If x = c is not at an endpoint, then f (c) is a local (relative) minimum value for f(x) if f(c) < or = f(x) for all x-values in a small open interval around x = c.
NECESSARY CONDITIONS FOR LOCAL EXTREME VALUES
If x is not at an endpoints, usually any local extrema will occur at x-values where the derivative f'(x) = 0 or f'(x) is undefined. This result is so important that you will automatically look at where derivative f'(x) = 0 or f'(x) is undefined if you are looking for local extrema. This necessary condition is stated as "Fermat's Theorem" or "The Local Extrema Theorem", which is Theorem 2 on page 247:
If a function f has a local (relative) extrema at x = c in an open interval, then either f'(x) = 0 or f'(x) is undefined.
The converse is not necessarily true. For example, you could have a function undefined at x = a and have neither local extrema because you have no derivative there. Or you could have the derivative equal to zero and have a horizontal asymptote at which you would have neither a local maximum or local minimum. Take a look ahead to Figure 4.23 on page 264 of your textbook. Points c1 and c5 have f'(x) = 0 but do not give local maximum or local minimum.
CRITICAL VALUES
These x-values where the derivative f'(x) = 0 or f'(x) are undefined are called critical values. The point where the derivative has these characteristics is called a critical point.
How will you find these critical values? First find the derivative of the function. Then set the derivative equal to zero and solve for any values of the independent variable that makes the derivative zero. Also look for any values of the independent variable where the derivative is undefined.
DETERMINE ABSOLUTE EXTREMA IN CLOSED INTERVAL
Now that we have learned about Fermat's theorem and critical values, we are in a position to determine absolute extrema values for a function defined in a closed interval. How do that?
- First evaluate the function f at all endpoints.
- Second find all critical values using Fermat's Theorem.
- Evaluate the function f at all critical values.
- If you have a closed interval, then the greatest of your evaluations is the absolute maximum value for the function f and the least value is the absolute minimum value for the function f.
In your textbook, Examples 2, 3, 4, and 5 on pages 249 - 252 illustrate this process. Make sure you work the problems on a separate sheet of paper without looking at the solutions in the examples.
Now you can go to the Session 12 assignments to continue your assignment.
At the end of this session you will be able to
STATE ROLLE'S THEOREM
Michael Rolle, a Frenchman, published this theorem in 1691 approximately 20 years after Isaac Newton's invention of differential calculus.
As you look at Rolle's Theorem on page 255, you noticed that there are three conditions in the hypothesis:
- Function f is continuous in the closed interval from a to b.
- Function f is differentiable in the open interval from a to b.
- f(a) = f(b) = 0.
Then the conclusion is that there is x = c in the open interval such that f'(c) = 0.
DESCRIBE WHAT THE THEOREM STATES
First, the statement that f(a) = f(b) = 0 tells us that the evaluation of the function at the end points is zero. Thus, the endpoints are on the horizontal axis.
Consequentially, Rolle's Theorem tells us that if a function is continuous on the closed interval, differentiable on the open interval, and the evaluation of the function at the end points is zero, then we know someplace in the open interval that the derivative is zero.
Because one application of the derivative is instantaneous rate of change, this theorem states that some place in the open interval the instantaneous rate of change of change is zero.
GEOMETRIC REPRESENTATION OF THE THEOREM
In the hypothesis of the theorem, continuity on the closed interval tells you that you can go on the curve from the left end to the right end along the curve.
In the hypothesis, differentiability on the open interval tells you that the derivative exists at every point in the open interval, you have a tangent line to the curve at every point, and the curve is smooth.
Last, in the hypothesis, the end points are on the horizontal axis.
In the conclusion, we know that there is at least one point in the open interval between the endpoints where the slope of the curve is zero and we have a horizontal tangent line because the derivative geometrically is the slope of the curve. Figure 4.10 on page 255 of the textbook shows three places where the conclusion holds.
Now you can go to the Session 12 assignments to continue your assignment.
At the end of this session you will be able to
THE MEAN VALUE THEOREM
As you look at Mean Value Theorem on page 257, you noticed that three are two conditions in the hypothesis:
- Function f is continuous in the closed interval from a to b.
- Function f is differentiable in the open interval from a to b.
Then the conclusion is that there is x = c in the open interval such that f '(c) = (f(b) - f(a)) / (b - a).
Do notice that the endpoints of the interval do not have to be on the horizontal axis or even have to be equal.
DESCRIBE WHAT THE THEOREM STATES
The Mean Value Theorem tells us that if a function is continuous on the closed interval and differentiable on the open interval, then we know there is a place in the open interval between the endpoints where the derivative is the same as (f(b) - f(a)) / (b - a).
Recalling that (f(b) - f(a)) / (b - a) is the average rate of change of the function and f'(c) is the instantaneous rate of change, you then see there is a point in the open interval between the endpoints where instantaneous rate of change of the function is exactly the same as average rate of change of the function between the endpoints.
GEOMETRIC REPRESENTATION OF THE THEOREM
In the hypothesis of the theorem, continuity on the closed interval tells you that you can go on the curve from the left end to the right end along the curve.
In the hypothesis, differentiability on the open interval tells you that the derivative exists at every point in the open interval, you have a tangent line to the curve at every point, and the curve is smooth.
(f(b) - f(a)) / (b - a) is the slope of the straight line, the secant line, joining the end points.
f'(c) is the slope of the curve, that is the slope of the tangent line at x = c.
Therefore, geometrically, if we have a continuous, smooth curve, then at a place in the open interval the slope to the curve is exactly the same as the slope of the straight line passing through the endpoints. In other words the tangent line is parallel to the straight line passing through the endpoints.
Now you can go to the Session 12 assignments to continue your assignment.
At the end of this session you will be able to
CONSEQUENCES OF THE MEAN VALUE THEOREM
The first consequence of the Mean Value Theorem is, if the derivative is zero every place in an open interval, then the function must be a constant throughout the open interval. Briefly, if f '(x) = 0 in an open interval, then f(x) is a constant in the open interval.
The second consequence of the Mean Value Theorem is, if the derivative of two different functions are equal in an open interval, then the functions differ by a constant value. Briefly, if f'(x) = g '(x) in an open interval, then f(x) - g(x) = constant or f(x) = g(x) + constant.
Thus the second consequence of the Mean Value Theorem is, that, if we know the function f '(x), there is not a unique f(x) that will give f'(x) when we differentiate. Any two f(x) you have will differ by a constant "C". Regardless the constant "C" contained in f(x), when you differentiate, the derivative of the "C" will be zero.
In both situations we usually write a capital "C" for the constant.
For example, if f(x) = sec x + 100, g(x) = sec x - 123 and h(x) = sec x + C, then they all have the same derivative sec x. So if you are given the derivative as sec x, you do not know which function gave you the derivative. Consequently, we usually write f(x) = sec x + C for the function with the understanding the C could be any real number constant.
Now you can go to the Session 12 assignments to continue your assignment.
At the end of this session you will be able to
DESCRIBE A DIFFERENTIAL EQUATION
A differential equation is an equation that relates an unknown function and one or more derivatives. In other words, a differential equation is an equation that has at least one derivative in it.
SOLVE AN ELEMENTARY DIFFERENTIAL EQUATION
To solve an elementary differential equation means to find a function that solves the differential equation, that is to find a function that, with it's derivatives, makes the differential equation true.
To find the function requires us to know the differentiation formulas. Also, to find the function requires us to find the values of "C". To find a values of C when we know that f(x) = g(x) + C, we substitute a pair of values x = a and f(x) = f(a) and solve for C. Then substitute the value of C into f(x) = g(x) + C.
The following illustrate finding possible functions given a derivative:
1. If y' = sec t tan t, then y = tan t + C because, if we differentiate y, then we have the given y'.
2. If y' = 6 t^2, then y = 2 t^3 + C because, if we differentiate y, then we have the given y'.
3. If y' = 32 t, then y = 16 t^2 + C because, if we differentiate y, then we have the given y'.
4. If y' = 3x^2 - 4x + 5, then y = x^3 - 2x^2 + 5x + C because, if we differentiate y, then we have the given y'.
Now you can go to the Session 12 assignments to continue your assignment.