3.5 Graphing Functions

Guidelines for studying and graphing a function: (a) Define the domain. (b) Are there Vertical asymptotes? Horizontal asymptotes? (c) Is this function odd or even? (d) Find the derivative. What are the critical points? Give the intervals on which the function is increasing and decreasing. (Make TABLE 1). (e) Specify location and value of every local min or max. Which, if any, of the extreme values are absolute? (f) Find the second the derivative. Study concavity. Are there inflection points? (Make TABLE 2) (g) Make a graph of the function. Label axes with important values.

Prcatice 1 Do a complete function study of: y = x4 – 4x3 Discuss the curve with respect to concavity, points of inflection, and local maxima and minima. Use this information to sketch the curve. Solution: If f (x) = x4 – 4x3, then f (x) = 4x3 – 12x2 = 4x2(x – 3) critical numbers: we set f (x) = 0 and obtain x = 0 and x = 3. (Make Table1) f (3) = –27

- 0 - 0 + Table 1: x -∞ 0 3 ∞ x -∞ 0 3 ∞ f’(x) f’(x) - 0 - 0 + f (x) - 0 - 0 + x -∞ 0 3 ∞ f’(x) - 0 - 0 + f (x) f (0) = 0 f (3) = –27 Local min

f  (x) = 12x2 – 24x = 12x(x – 2) f (x) = 0 for x = 0 or 2 Table 2: ((Make Table2)) x -∞ 0 3 ∞ f (x) x -∞ 0 2 ∞ f’’(x) + 0 - 0 + The point (0, 0) is an inflection point since the curve changes from concave upward to concave downward there. Also (2, –16) is an inflection point since the curve changes from concave downward to concave upward there.

– Solution cont’d Using the local minimum, the intervals of concavity, and the inflection points, we sketch the curve.

Practice 2: Do a complete function study, follow the guidelines in the first slide. (Done in class).

Practice 3 Do a complete study of the function f (x) = x2/3(6 – x)1/3 Solution: Calculation of the first two derivatives gives Since f (x) = 0 when x = 4 and f  (x) does not exist when x = 0 or x = 6, the critical numbers are 0, 4 and 6.

f(0) = 0 is a local minimum. f(4) = 25/3 is a local maximum. The sign of f  does not change at 6, so there is no minimum or maximum there.

Looking at the expression f  (x): So f is concave downward on ( , 0) and (0, 6) concave upward on (6, ), and the only inflection point is (6, 0).

– Solution Graph: cont’d Note that the curve has vertical tangents at (0,0) and (6,0) because |f  (x)| as x 0 and as x 6.

Optimization Problems 3.7 Optimization Problems

An airline regulation says that all luggage that is box shaped must have the sum of height, width, and length not exceeding 64 in. What are the dimensions and volume of a square based box with the greatest possible volume under this rule?

Walking and swimming: You want to get from a “Start” point to a “Finish” point diametrically opposite each other around a circular pond with radius 1 mile. You first swim to point P at 2mi/hr then walk around the pond to the Finish point at 3mi/hr. Where should point P be so that you minimize your time of travel?

Ladder over fence: On the picture to the right, the fence is 8ft tall and is 3 ft away from the house. What is the shortest possible ladder that can clear the fence and still reach the house? ANSWER

3.9 Antiderivatives

Practice 1 Solution:

Prcatice 2 Solution: By extending the sum and difference rules to more than two terms, we get

Practice 3 Solution: First rewrite the integrand then work the integral.