Case 1: If for all then for all. Case 2: Since is a continuous function over the closed, bounded interval by the extreme value theorem, it has an absolute maximum. Also, since there is a point such that the absolute maximum is greater than Therefore, the absolute maximum does not occur at either endpoint. Case 3: The case when there exists a point such that is analogous to case 2, with maximum replaced by minimum. If is not differentiable, even at a single point, the result may not hold.
For example, the function is continuous over and but for any as shown in the following figure. Find all values where. The Mean Value Theorem states that if is continuous over the closed interval and differentiable over the open interval then there exists a point such that the tangent line to the graph of at is parallel to the secant line connecting and.
Let be continuous over the closed interval and differentiable over the open interval Then, there exists at least one point such that. Consider the line connecting and Since the slope of that line is. Let denote the vertical difference between the point and the point on that line. Consequently, there exists a point such that Since. Since we conclude that. In the next example, we show how the Mean Value Theorem can be applied to the function over the interval The method is the same for other functions, although sometimes with more interesting consequences.
For over the interval show that satisfies the hypothesis of the Mean Value Theorem, and therefore there exists at least one value such that is equal to the slope of the line connecting and Find these values guaranteed by the Mean Value Theorem.
We know that is continuous over and differentiable over Therefore, satisfies the hypotheses of the Mean Value Theorem, and there must exist at least one value such that is equal to the slope of the line connecting and Figure. To determine which value s of are guaranteed, first calculate the derivative of The derivative The slope of the line connecting and is given by.
We want to find such that That is, we want to find such that. Solving this equation for we obtain At this point, the slope of the tangent line equals the slope of the line joining the endpoints.
One application that helps illustrate the Mean Value Theorem involves velocity. For example, suppose we drive a car for 1 h down a straight road with an average velocity of 45 mph.
Let and denote the position and velocity of the car, respectively, for h. Assuming that the position function is differentiable, we can apply the Mean Value Theorem to conclude that, at some time the speed of the car was exactly.
If a rock is dropped from a height of ft, its position seconds after it is dropped until it hits the ground is given by the function.
Suppose a ball is dropped from a height of ft. Its position at time is Find the time when the instantaneous velocity of the ball equals its average velocity. First, determine how long it takes for the ball to hit the ground. Then, find the average velocity of the ball from the time it is dropped until it hits the ground. These results have important consequences, which we use in upcoming sections. At this point, we know the derivative of any constant function is zero.
The Mean Value Theorem allows us to conclude that the converse is also true. In particular, if for all in some interval then is constant over that interval. This result may seem intuitively obvious, but it has important implications that are not obvious, and we discuss them shortly. Let be differentiable over an interval If for all then constant for all.
Since is differentiable over must be continuous over Suppose is not constant for all in Then there exist where and Choose the notation so that Therefore,. Since is a differentiable function, by the Mean Value Theorem, there exists such that. Therefore, there exists such that which contradicts the assumption that for all. From Figure , it follows that if two functions have the same derivative, they differ by, at most, a constant. If and are differentiable over an interval and for all then for some constant.
Let Then, for all By Corollary 1, there is a constant such that for all Therefore, for all. The third corollary of the Mean Value Theorem discusses when a function is increasing and when it is decreasing. Recall that a function is increasing over if whenever whereas is decreasing over if whenever Using the Mean Value Theorem, we can show that if the derivative of a function is positive, then the function is increasing; if the derivative is negative, then the function is decreasing Figure.
We make use of this fact in the next section, where we show how to use the derivative of a function to locate local maximum and minimum values of the function, and how to determine the shape of the graph. This fact is important because it means that for a given function if there exists a function such that then, the only other functions that have a derivative equal to are for some constant We discuss this result in more detail later in the chapter.
Let be continuous over the closed interval and differentiable over the open interval. We will prove i. How is the average velocity of a moving object connected to the values of its position function? How do we interpret the average velocity of an object geometrically on the graph of its position function? Calculus can be viewed broadly as the study of change.
We begin with a simple problem: a ball is tossed straight up in the air. How is the ball moving? Questions like this one are central to our study of differential calculus. What does this value tell us about the motion of the ball? Any moving object has a position that can be considered a function of time.
On any time interval, a moving object also has an average velocity. For example, to compute a car's average velocity we divide the number of miles traveled by the time elapsed, which gives the velocity in miles per hour. On the graph provided in Figure 1. In light of this meaning, what is a geometric way to interpret each of the values computed in the preceding question? Whether we are driving a car, riding a bike, or throwing a ball, we have an intuitive sense that a moving object has a velocity at any given moment -- a number that measures how fast the object is moving right now.
Log in. A position function is provided, wheres represents miles and t represents hours. Find the average velocity on the four intervals provided, then estimate the instantaneous velocity at the time that begins each interval. So we're gonna start with our first time interval from seconds. So we need to find s of three like three into the function you'll get 46 And as of four you plug four into the function you'll get 1 And that gives us Now we're gonna keep using this very first one.
So I put a little star beside that. So let's go to our next time interval which is between three and 3. So we need to find our function at 3. Since we already know the function at three and 35 it is 77 75 So we do our slope formula again And that gives us 63 5 for that slow. Then our next time interval between three and three point once we have to find Our function at 3.
And we do our slope formula again And it gives us 55 And then our final time interval 1. We have to find s of 31 and that is 46 We do our slope formula one last time And that gives us 54 to.
So we can see as our time intervals got closer together In our last time interval here is only 31 at time three uh are instantaneous velocity. We can make an estimate that's somewhere I'd say close to 54 to because you can see these we get close.
In mathematics, precalculus is the study of functions as opposed to calculus, which is the study of change, and algebra, which is the study of operations and their application to solving equations.
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