Definite Integrals Solutions to Practice Problems for AP Calculus
Practice problems for these solutions can be found at: Definite Integrals Practice Problems for AP Calculus
Part A The use of a calculator is not allowed.
Note that f (x )= cos x – x2 is an even function. Thus you could have written and obtained the same result.
Part B Calculators are allowed.
- See Figure 11.9-1.
- See Figure 11.9-2.
- A change of concavity occurs at x = – 3, x =1, and x =5.
The function f has a relative minimum at x = – 5 and x =3, and f has a relative maximum at x = – 1 and x =7.
The function f is concave upward on intervals (–6, –3) and (1, 5).
Thus at each of the points at (1, 1) and (1, –5) the graph has a horizontal tangent.
Enter [Solve] (9x 2 – 8x – 27=0, x ) and obtain x =3 or x = – 1.
Thus at each of the points (3, –2) and (–1, –2), the graph has a vertical tangent. See Figure 11.9-3.
Step 1. See Figure 11.9-4. Let P =x + y where P is the length of the pipe and x and y are as shown. The minimum value of P is the maximum length of the pipe to be able to turn in the corner. By similar triangles,
Step 2.Find the minimum value of P. Enter Use the [Minimum] function of the calculator and obtain the minimum point (9.306, 22.388).
Step 3.Verify with the First Derivative Test. Enter y 2=(y1(x ), x ) and observe. (See Figure 11.9-5.)
Step 4.Check endpoints. The domain of x is (6,∞). Since x =9.306 is the only relative extremum, it is The absolute minimum. Thus the maximum length of the pipe is 22.388 feet.
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