The quiz problem was to determine first the location of any maxima or minima for the function f(t) = 3 t^2 - 4t + 8. The derivative this function is f'(t) = 6t - 4, with resulting critical point t = 2/3.
The first derivative test tells us that 6 t - 4 > 0 when t > 2/3, and 6t - 4 < 0 when t < 2/3. The slope of the graph therefore goes from negative to positive at the critical point t = 2/3. This implies that there is a minimum at this critical point.
We could also have used the second derivative test to determine that we have a minimum at the critical point. The second derivative of the function is f''(t) = 6, which is always positive. This implies that the graph of the function is everywhere concave upward and that any critical point will therefore give us a minimum.
We now extend our investigation to the function family defined by f(t) = 3 t^2 + b t + 8. The derivative is f'(t) = 6t + 8, and the second derivative as before it is f''(t) = 6.
We see that the critical point is at t = -b / 6, and that since the second derivative is positive this critical point will give us a minimum.
It is interesting to locate the graph position of the minimum point. When t = -b / 6, the corresponding y value will be y = f(-b/6) = 3(-b/6)^2 + b(-b/6) + 8 = ... = - b^2 / 12 + 8. The minimum point therefore occurs at (-b/6, -b^2/12 + 8).
We can express the coordinates of these minimum points for various functions in the family in terms of their t values. A minimum point occurs at coordinates (t, y) = (-b/6, -b^2/12 + 8). Thus we have t = -b/6 and y = -b^2/12 + 8. We can eliminate b from these equations; solving the first we get b = -6t, and substituting this into the second we obtain y = - (-6t) ^ 2 / 12 + 8 = -3t^2 + 8.
We therefore conclude that for any value of t, there will be a value of b such that the minimum point for the function f(t) = 3t^2 + bt + 8 occurs at the point (t, -3t^2 + 8).
It follows that if we sketch a graph of y = -3t^2 + 8 vs. t, at every point of this graph there will be a parabola representing the function f(t) = 3t^2 + bt + 8, with its minimum point on the graph. This situation is depicted rather poorly by the graph below (the tiny parabolas with their minimum points on y = -3t^2 + 8 should spread out just as fast as to graph of that function).
A DERIVE representation of the situation confirms these results in a rather striking way. We can easily alter and sketch the graph of the function y = -3 t ^ 2 + 8. We can then use various values of b to author functions of the family f(t) = 3 t^2 + b t + 8.
A couple of DERIVE tricks (they are really tricks, but rather standard procedures; but it's more fun to treat them as tricks) can facilitate this process. If we author the function f(b, t) := 3 t^2 + b t + 8 (we use the := to assign the expression to the function; otherwise DERIVE will think we're trying to multiply f by (b; t), whenever that might mean), then the expression f(2, t) will stand for the function 3 t^2 + 2 t + 8; we can highlight the expression f(2,t) and plotted and we will get a plot of this specific function. This is easier than authoring a bunch of different functions in the family.
Whenever we do to author our functions, we see that when we plot them every function in the family will have a vertex on the plot of the 'vertex locator' function f(t) = -3 t^2 + 8.
If you do this you will be impressed. But it gets better. If we have defined the function f(b,t) := 3 t^2 + b t + 8, we can then author the expression VECTOR(f(b,t), b, -4, 4, .25). If we simplify this expression, we will get all 32 of the functions f(t) = 3 t^2 + b t + 8 corresponding to the b values -4, -3.75, -3.5, . . . , 3.5, 3.75, 4. Not only that, but if we plot this resulting expression DERIVE will plot all 32 graphs. Try it. You will like it.
If you are and on-campus student, these commands are contained in the DERIVE file d:/fall98/ca1request for this file and it will be sent as an attachment. You can Transfer Load the file and see the precise syntax of each command; you can also of course plot the various expressions.
We now look at the function family y = x - k `sqrt(x). We are told that the minimum for this function always lies 1/4 of the way between its zeros (express below as halfway to halfway to correct a fairly obvious error made by the instructor).
To verify this claim, we must find the location of the minimum point for a general function of this family, and we must also find its zeros.
The derivative is easily verified to be y' = 1 - k / (2 `sqrt(x)), x > 0; setting this expression equal to zero we find that the critical point for the function is at x = k^2 / 4.
The second derivative of the function is easily verified to be k / 4 x^(3/2) (note error in the figure below; the 1 will not appear in the second derivative, for reasons that should be obvious). This second derivative will clearly be positive for positive values of x. It follows that the critical point will give us a maximum for the function.
To verify that the critical point lies 1/4 of the way between the zeros of the function, we need to find zeros of the function. This is easily done by solving the equation x - k `sqrt(x) = 0. We factor `sqrt(x) out of the left-hand side, obtaining factors `sqrt(x) and `sqrt(x) - k; the solutions to the equation can then be easily found. We obtain zeros at x = 0 and at x = k^2.
Since the zeros are at x = 0 and x = k^2, we see that the critical point x = k^2 / 4 is in fact 1/4 of the way between the zeros.
We can use DERIVE to illustrate our conclusions. Using the techniques described above, we can easily enough plot a number of members of the function family and visually verify that the critical point, and therefore the minimum value of each, lies at index value that is 1/4 of the way between the zeros.
We next look at the important family y = a x e^(-bx). We wish to locate its relative maximum and minimum points.
We begin by taking the derivative of this function to find its critical point(s). We factor out the a and use the product rule, then the quotient rule, and we finally factor out the common e^-(bx) to obtain y' = a e^(-bx) (1-bx).
Setting this expression for y' equal to 0, we can divide by a e^-(bx), which can never be 0 (otherwise we couldn't divide by it without noting the exceptions when expression was 0), and easily solving resulting equation to see that the only critical point occurs at x = 1/b.
We can take the derivative of y' to obtain y'' = a b e^-(bx) (bx - 2), which when evaluated for the critical value x = 1/b gives us y''(1/b) = -a b e^-1. Since a and b are assumed to be positive, and since e^-1 is positive we see that the second derivative is negative. It follows that the function has a maximum at its critical point x = 1/b.
We therefore expect that if we graph the family of functions for increasing values of b, that x coordinates of the maxima will decrease and the maxima move to the left with increasing b.
To determine whether the current function has a point of inflection, we find where the second derivative is positive, corresponding to upward concavity, and where it is negative with downward resulting concavity.
Solving the inequality a b e^-(bx) (bx - 2) < 0 for x, we first note that the factor a b e^-(bx) is always positive so that we are safe in dividing both sides of the equation by this quantity (if the quantity was negative we would have to reverse the direction of the inequality). We easily solve the resulting inequality to obtain x < 2/b, indicating that the graph is concave downward whenever x <2/b.
We solve the inequality corresponding to y'' > 0 in almost exactly the same way, reaching the conclusion that the graph is concave upward whenever x > 2/b.
It follows that the point x = 2/b, where the graph goes from concave downward to concave upward, is the point of inflection for the function.
We should note that the location of the critical point at x = 1/b is halfway as far along the x axes as the location of the point of inflection at x = 2/b. You should use DERIVE to graph several members of the family and verify these properties of the graph.
We note that when a function is defined over an interval, it can have relative maxima inside the interval, and that one of these relative maxima might well be the maximum of the function over the interval. However, it is also possible that the maximum of the function over the interval occurs at one of the endpoints of the interval. Therefore to find a relative maximum or minimum, we have to first find the value of the greatest of the relative maxima and the least of the relative minimum; we then have to compare these values with it to endpoint values to be sure that they really 'hold up' as the maximum and minimum over the interval.
We call the maximum and the minimum over the entire interval the global maximum and the global minimum.
In the figure below we see that the greatest of the relative maxima is less than the value of the function at the right-hand endpoint of its interval of definition. The global maximum therefore occurs at this endpoint, even though it is not a relative maximum. The global minimum, however, occurs at the indicated relative minimum, at which point the value of the function is less than that of any other relative minimum, and is also less than the value of the function at either endpoint.
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