Pre-Calculus Test 3

Instructor: Prof. A. Mingarelli
Mar. 1997

NOTATION: Pi = 3.14159... is our usual Greek number

1. Let f(x) = 2 + 5x - 12 x2. Find the intervals where the graph of f is (a) concave up and concave down, (b) increasing and decreasing and where (c) the graph has critical points and points of inflection.

Solution: Note that f'(x) = 5 - 24x; f"(x) = -24. Thus, f is increasing when 5 - 24x > 0, i.e., x < 5/24 and f is decreasing when 5 - 24x < 0, i.e., x > 5/24. Critical point occurs at x = 5/24. Point of inflection: None, because f"(x) is never equal to zero.

2. Find the horizontal and vertical asymptotes of the graph of the function f(x) = x2/sqrt(x+1).

Solution: Vertical asymptote occurs when x + 1 = 0, i.e., when x = -1. For horizontal aymptotes we need to calculate the limit as x -> +/- infinity of the function f: It is not difficult to see that x2/sqrt(x+1) = x2/{sqrt(x) sqrt(1 + 1/x)} = x3/2/sqrt(1 + 1/x). So, as x -> infinity the last term approaches infinity as well, and there are no (finite) horizontal aymptotes. As x -> - infinity the function f is not even defined and so there cannot be a horizontal asymptote there.

3. How fast is the area of a square increasing when the side is 3m. in length and growing at a rate of 0.8 m/min.?

Solution: Let x = x(t) denote the side of the square. Its area A(x) = x2, and by the Chain Rule we know that A'(x) = 2 x(t) x'(t) = (2)(3)(0.8) = 4.8 m2/min.

4. A box with a square base and open top must have a volume of 4000 cm3. Find the dimensions of the box that minimize the amount of material used.

Solution: Let x denote the sides of the square base of the box. and let y denote its height. Then we are given that the volume, namely x2y = 4000. We want to minimize the surface area, A, of the box, right? Now, A is given by A = x2 + 4xy = x2 + 4x(4000/x2) = x2 + 16000/x. We need to find the critical point(s) of f...So, we set f'(x) = 0 where f'(x) = 2x - 16000/x2 = 0. This gives x = 20 as the other roots are not real numbers. Notice that A"(20) > 0 and so, by the second derivative test, x = 20 provides a minimum for the Area, A. The dimensions are therefore x = 20, x = 20, (for the base) and y = 10 (for the height).