AP Exam Practice Quiz 1 and Intro to the Intermediate Value Theorem of Integrals

Introducing the first of a series of AP exam practice quizzes, the intermediate value theorem of integrals, the day when the AP Calculus 2008: Without Bound blog is unblocked at Daniel McIntyre Collegiate Institute, the series of YouTube videos in tandem with Flickr pics, and kristina's powerpoint presentations!

AP EXAM PRACTICE QUIZ 1


Let h be a function defined for all x does not equal 0 such that h(4) = -3 and the derivative of h is given by h'(x) = (x^2-2)/x for all x does not equal 0.


a) Find all values of x for which the graph of h has a horizontal tangent, and determine whether h has a local maximum, a local minimum, or neither at each of these values.

To determine if there are local extrema at the critical numbers, we must first determine the critical numbers. Remember that there are critical numbers when a function has an asymptote or is undefined. h' = 0 when x = sqrt(2) and -sqrt(2). h' is undefined at x = 0 since 0 isn't in the domain (as stated in the question). So critical numbers are x = sqrt(2), -sqrt(2), and 0.

We use the first derivative test on h' to see where h' is positive or negative. Why? Because when h' is positive, h is increasing; when h' is negative, h' is decreasing. So a change in sign in h' would indicate a slope of zero at that point and that's where there are local extrema. According to the line analysis, we see that to the left of -sqrt(2), h' is negative; between -sqrt(2) and 0, h' is positive; between 0 and sqrt(2), h' is negative; and to the right of sqrt(2), h' is positive. Wherever h' changes sign from negative to positive, h has a local minimum; wherever h' changes sign from positive to negative, h has a local maximum. By the first derivative test (line analysis), there are local minimums at x = -sqrt(2) and x = sqrt(2). We don't look at 0 because it's not part of the domain of the function.


b) On what intervals, if any, is the graph of h concave up?

Rememberize these rules (from chapter 5 of your textbook):

If the second derivative is positive, the first derivative is increasing, and the parent function is concave up.

If the second derivative is negative, the first derivative is decreasing, and the parent function is concave down.

So wherever h" is positive, h is concave up.

We determine h" by differentiating h using the quotient rule.

We see that h" is positive, so h is concave up everywhere.


c) Write an equation for the line tangent to the graph of h at x = 4.

Pull out the point-slope formula: y-y1=m(x-x1)

The question gave us the x-coordinate: x = 4.
The question gave us the y-coordinate: y= -3.
m is the slope at x = 4, so plug x = 4 into h' which spits out 7/2.

Plug those numbers into the equation. BING! BANG! BOOM! We're done part c.

y+3=(7/2)(x-4)


d) Does the line tangent to the graph of h at x = 4 lie above or below the graph of h for x > 4?

If we draw a line tangent to the graph at x = 4, the line is below the graph at x > 4, because h is concave up everywhere.


INTERMEDIATE VALUE THEOREM OF INTEGRALS

Similar to the intermediate (or mean) value theorem of derivatives (that in a closed interval between a and b there exists a point on a continuous function which equals the average value), there exists a point on a continuous function which equals the average integral of the function.


We have b = 3. We have a = 0. We have f(x) = 1-2x. Plug the numbers into the equation. BING! BANG! BOOM! We get the average integral. But why does it work?

In this graph, pivot the yellow area found between c and b to the white area found between c and a. Notice that they fit together like a jigsaw puzzle to yield one big yellow rectangle. We can imagine that one big rectangle having the same area as the area under the graph, which we can see as the average value of the function.

HOUSEKEEPING

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