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course MTH 174
1/28 5:40PM
If your solution to stated problem does not match the given solution, you should self-critique per instructions at
http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm
.
Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.
The derivative of y = f(x) with respect to x is the instantaneous rate of change of y with respect to x.
The definite integral of a function y = f(x) over an interval between x = a and x = b is equal to the change in an antiderivative function between x = a and x = b over that interval.
The derivative function is the ‘rate-of-change function’.
An antiderivative function is the ‘change-in-quantity function’, in the sense that if y = f(x) represents the rate of change of some quantity Q with respect to x, then the change in the quantity Q over an interval is equal to the change in the antiderivative function over that interval.
More complete statements:
The derivative function f ‘ (x) of a function y = f(x), with respect to variable x, is the instantaneous rate of change of y with respect to x. The rate of change of y with respect to x at a specific point x = x0 is f ‘ (x0).
The change in the function y = f(x) between specific points x = a and x = b is equal to the change in an antiderivative function F(x) between x = a and x = b, provided an antiderivative function exists on this entire interval. When this is the case, the change in the antiderivative function
An antiderivative function F(x) is any function F(x) whose derivative is equal to f(x).
Antiderivative functions are known up to an arbitrary integration constant so if an antiderivative function exists on an interval, there are infinitely many antiderivative functions on that interval.
To find the value of a definite integral, one specific antiderivative function must be chosen.
The change in the chosen antiderivative function is (final value) - (initial value) = F(b) - F(a).
For a given function f(x) on a given interval:
the derivative may be defined everywhere on that interval
the derivative may be defined nowhere on that interval
the derivative may be defined at all but a finite number of points
the derivative may be defined at all but an infinite number of points (either countably infinite or uncountably infinite).
Nearly all of the functions studied in first-year calculus, and most of the functions required to apply calculus to the real world, have derivatives defined on easily-recognized intervals.
For many of the functions studied in first-year calculus, the derivative is defined for all real numbers.
For some functions there are specific points or intervals where the derivative is not defined, but between these points and intervals are the easily determined intervals on which the derivative is defined.
The set of intervals (and/or points) over which a derivative function is defined is the domain of the derivative function.
If the derivative is defined at all points of an interval, then the function is said to be differentiable on that interval, and the derivative function f ‘ (x) will then give the rate of change of y with respect to x at all points of the interval.
If a function f(x) has an antiderivative F(x) at every point of an interval, then it is said to be integrable over that interval.
If this is the case, then between any two points of the interval, the change in the antiderivative gives us the definite integral of f(x), with respect to x, between those points.
Note that for f(x) to have an antiderivative F(x) at every point of an interval, F(x) must be differentiable at every point of the interval.
So if there is a point (or an interval) between x = a and x = b where the derivative does not exist, the change in an antiderivative function cannot be expected to give us the value of the definite integral; in fact the definite integral would not be defined at all for this interval.
Any function f(x) that can be written down as a series of constant multiples, sums, products, quotients and composites of the basic power, exponential, logarithmic and polynomial functions studied in first-year calculus has a derivative f ' (x) that can be found by correctly using the constant-multiple, sum, product, quotient and chain rules, and the expressions for the derivatives of these basic functions. While the calculations can be messy, they aren't tricky. If the method is followed correctly, it works.
However even though an antiderivative F(x) of a such a function f(x) might exist, it is not always possible to express the antiderivative in terms of the basic functions. When it is possible, it is often tricky, and there's really no limit to how tricky it can get. In second-semester calculus we learn some of the basic tricks, and we learn how to approximate definite integrals when the tricks don't work.
An important basic picture:
If y(t) is the depth of water in a container, then the derivative y ' (t) is the function which gives you the rate of change at any instant.
If r(t) is the rate at which depth is changing, then r(t) = y ' (t). If all you know is r(t) then it follows that y(t) is an antiderivative of r(t).
The change in y between any two t values t = t1 and t = t2 is therefore equal to the change in the antiderivative between those t values: change in y = antiderivative of r(t) evaluated at t2 - antiderivative of r(t) evaluated at t1. That is, the change in y is the definite integral of the rate function from t = t1 to t = t2.
In this example we see a clear illustration of the Fundamental Theorem of Calculus.
Calculus II
Asst # 1
07-13-2001
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Optional preliminary questions:
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Question: `q001. The domain of f(x) is the interval 1 <= x <= 5. Its graph is a straight line from (1, 3) to (5, 5). Sketch the graph of f(x), its derivative function f ' (x), and the function F(x) whose value at x = 1 is 7 and whose derivative F ' (x) is equal to f(x).
Answer the following without symbolically integrating or differentiating f(x):
Describe the graph of its derivative function f ' (x) on this interval.
Describe the graph of its antiderivative function F(x), with the initial condition F(1) = 7.
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Your solution:
confidence rating #$&*:
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Question: `q002. The domain of g(x) is the interval 1 <= x <= 8. Its graph consists of a straight line from (1, 3) to (5, 5) (so that it coincides with the graph of the function f(x) of the preceding problem on the interval for which f(x) is defined) and then along another straight line from (5, 5) to (8, 12). Sketch the graph of g(x), its derivative function g ' (x), and the function G(x) whose value at x = 1 is 7 and whose derivative G ' (x) is equal to g(x).
Describe the graph of its derivative function g ' (x) on this interval.
Describe the graph of its antiderivative function G(x), with the initial condition G(1) = 2.
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Your solution:
confidence rating #$&*:
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Question: `q003. The domain of h(x) is 0 <= x <= 6. Its graph consists of a smooth curve passing through the points (1, 3), (4, 5) and (6, 1), with its peak at (4, 5). The graph increases at a decreasing rate between x = 1 and x = 4, and decreases at an increasing rate between (4, 5) and (6, 1).
Sketch the graph of h(x), its derivative function h ' (x), and the function H(x) whose value at x = 1 is 7 and whose derivative H ' (x) is equal to h(x).
Based on your graph of h(x), what is your best estimate for the maximum value of h ' (x) on the interval 0 <= x <= 6? What is your best estimate for the minimum value of h ' (x) on this interval? What are the coordinates of the point at which h ' (x) crosses the x axis, if indeed it does so?
Based on your graph of h(x), what is your best estimate for the change in the value of H(x) between x = 1 and x = 4, and for the change in the value of H(x) between x = 4 and x = 6?
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Your solution:
confidence rating #$&*:
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Self-critique Rating: