Master AP Calculus AB & BC
Part II. AP CALCULUS AB & BC REVIEW
CHAPTER 2. Calculus Prerequisites
TECHNOLOGY: SOLVING EQUATIONS WITH A GRAPHING CALCULATOR
You know how to solve equations; you’ve been doing it successfully since introductory algebra, and you’re not even afraid of quadratic functions. “Bring ‘em on,’’ you say, with a menacing glint in your eye and a quadratic formula program humming in your calculator’s memory. One problem: the College Board also knows that you have access to calculator technology, and some of the equations you’ll be asked to solve aren’t going to be the pretty little factorable ones you’re used to. In fact, some equations on the calculator-active section will look downright ugly and frightening.
The College Board expects that you will have access to and proficiency with a graphing calculator that can do four major things, at the least, as described in Chapter 1. You must be able to solve equations (even ugly ones) with your graphing calculator. As always, instructions are included for the TI-83, currently the most common calculator used on the AP test. Consult your instruction manual if you have a calculator other than the TI-83.
Example 24: Find the solutions to the equation e2x + cos x = sin x on [—2π,2π].
Solution: There is no easy way to solve this; it’s just enough to shake your faith in the usefulness of the quadratic formula. However, there is hope! First, move things around so that the equation equals zero. If you subtract sin x from both sides of the equation, that goal is accomplished: e2x + cos x — sin x = 0. Now, the solutions to the original equation will be the roots of this new equation. Put your calculator in radians mode ([Mode] → “Radian”) and graph Y = e2x + cos x — sin x. If you press [Zoom] → “Ztrig”, the window is nicely suited to trigonometric functions—the window is basically [—2π,2π] for the x-axis and [—4,4] for the y-axis. If you did everything correctly, you should see this:
Finding those x-intercepts is our goal. It doesn’t matter which you find first, but we’ll start with the rightmost one. Clearly, the root falls between x = —π and —π/2. To calculate the root, press [2nd] → [Trace] → “zero,” as we are looking for zeros of the function. At the prompt “Left bound?” type a number to the left of this root; for example, —π. Similarly, for the prompt “Right bound?” you can type 0.
At the prompt “Guess?” you should take a stab at prognosticating the root. A good guess seems to be —3π/4. Once you press [Enter], the calculator does the rest of the work for you, and the root turns out to be x = —2.362467. The AP test only requests three-decimal place accuracy, so —2.362 is acceptable. Note, however, that 2.363 is not correct and is not accepted. You do not have to round answers—you can simply cut off (or truncate) decimals after the thousandths place.
Follow the same steps to get the second root. If you don’t like typing in guesses for the boundaries, you can press the left and right arrow buttons to move the little “X” turtle along the graph. For example, in the diagrams below, you move the little turtle to the left of the root at the prompt “Left bound?”. Once you press Enter and get the “Right bound?” prompt, you move the turtle to the right of the root. You can even do this when asked for a guess—just move the turtle between the two boundaries and get close to the root.
The second root of the equation is —5.497775, which can be written as —5.497 or —5.498 on the AP test.
Directions: Solve each of the following problems. Decide which is the best of the choices given and indicate your responses in the book.
YOU MAY USE A GRAPHING CALCULATOR FOR PROBLEMS 9 AND 10 ONLY.
For problems 1 through 5, use the following graphs of f and g:
1. Determine which of the two functions has an inverse and sketch it.
2. Graph |f(x)| and g(|x|).
3. If/is created with a semicircle and an absolute value graph, write the equation that represents f(x).
4. Draw a function h(x) such that h(x) = g-1(x) when x ≤ 0 and h(x) is odd.
5. Evaluate f(g(f)-2))).
6. Solve the equation cos 2x — cos2x = 2sin x and give solutions on the interval [0,2π).
7. If m(x) = 2x3 + 5x — 2, find m-1(6).
ALERT! A bad choice for the bounds would have been [—3π,0], as this interval includes both roots. When you are trying to calculate a root, make sure only the root you are trying to find falls with in the boundaries you assign.
*8. Graph r = cos θ — sin θcos θ and find the values of θ where the graph intersects the pole.
*9. Graph x = et, y = t + 1, and express the parametric equations in rectangular form.
*10. James’ Diabolical Challenge: Write a set of parametric equations such that: at t = 0, x = 0 and y = 0 and at t = 4, x = 4 and y = 6. Then, find the rectangular inverse of your equations.
*BC —only problem
ANSWERS AND EXPLANATIONS
1. Only g has an inverse function; because f fails the horizontal line test, it is not one-to-one and thus has no inverse. To graph g-1, either reflect g about the line y = x or choose some coordinates from the graph of g and reverse them—remember that if (a,b) is on the graph of g, then (b,a) is on the graph of g-1.
2. The graph of |f(x)| will not extend below the x-axis, and g(|x|) will be y-symmetric, as shown below:
3. The circle has equation (x + 2)2 + y2 = 4, so solving for y and identifying only the upper half of the circle results in the equation The absolute value graph is y = |x — 1| — 1, by graph translations. Therefore, we can write the multi-rule function.
4. The graph of h will look exactly like the graph of g-1 for x ≤ 0 (as the functions are equal there), but since h is odd, its graph will have to be origin-symmetric, which dictates the graph of h for x > 0.
5. From the graphs, you see that f(—2) = 2, g(2) = 1, and f(1) = —1. Thus, f(g(f(—2))) = —1.
6. Using a double-angle formula for cos 2x, rewrite the equation to get cos2x — sin2x — cos2x = 2sin x. This simplifies easily to —sin2x — 2sin x = 0. Now, factor the equation: —sin x(sin x + 2) = 0. The answer is x = 0, π (remember that you cannot solve sin x = —2, since sine has a range of 0 ≤ y ≤ 1).
7. First of all, if 6 is in the domain of m-1, then it is in the range of m. Therefore, there is some number x such that 2x3 + 5x — 2 = 6. It is not easy to find that number, however, and you should resort to the graphing calculator to solve the equivalent equation 2x3 + 5x — 8 = 0 in the method described in the Technology section of this chapter. Doing so results in x = 1.087. Thus, m (1.087) = 6 and m-1(6) = 1.087.
8. The graph, given below, will hit the pole whenever r = 0. Therefore, set the equation equal to 0 and factor to get cos θ(1 — sin θ) = 0. This has solutions θ = π/2 and 3π/2.
9. The graph is given below. In order to put in rectangular form, solve x = et for t by taking ln of both sides. Doing so results in ln x = t. (You could also solve the y equation for t, but this results in a much uglier rectangular form—a function of y.) Substituting t = ln x into the y equation gives y = ln x + 1, x > 0. The restriction is caused by the range of et being positive, as denoted earlier in the chapter.
10. The simplest solution for this problem is x = t and y = 3/2t, as 6 is 3/2 of 4. The rectangular form of this problem is y = 3/2x, so the inverse is y = 3/2x.
SUMMING IT UP
• Calculus is rife with functions so it’s important that you understand what they are: a function is a special type of relation.
• If a function does not fulfill the requirements to be even or odd, the function is classified as neither even nor odd.
• It is easy to remember that origin-symmetric functions are odd—they both start with the letter “o.”
• Plugging one function into another is called composing the function with another.
• Remember not to stress out over exact graphs.
• Radians are used almost exclusively on the AP test (in lieu of degrees).
• Some students record unit circle values in their calculator’s memory instead of memorizing them. Remember that approximately 50 percent of the AP test is taken without a calculator. Make sure to memorize the unit circle!
• When giving solutions to inverse trigonometric function, always remember the bubble.