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$100 What is the derivative of (3x 2 ) 1/2 ? B - (1/2)(6x) -1/2 A - (1/2)(3x 2 ) -1/2 C - (6x) / 2x(3) 1/2 D - (2/3)(3x 2 ) 3/2 c - (6x) / 2x(3) 1/2

Explanation Y = (3x 2 ) 1/2 d/dx[u n ] = nu n-1 u’ u = 3x 2 n = 1/2 So, d/dx (3x2)1/2 = 1/2(3x2) -1/2 (6x) = (6x) / 2x(3) 1/2

$200 How many critical numbers are on the graph of 2x 2 (4x) B - 2 A - 1 C - 3 D - 4 A - 1

Explanation Critical numbers exist where the f’(x) = 0 f(x) = 2x 2 (4x) or 8x 3 So f’(x) = 24x 2 24x 2 = 0 Divide each side by 24 to get x 2 = 0 Square root each side to find that x = 0 and that there is only one critical number.

$500 If mean value theorem applies, find all values of c in the open interval (a,b) such that f(x) = x 2/3 [0,1] B - c = A - c = C - c =.296 D - Mean value does not apply C - c =.296

Explanation Mean value applies because f(x) is differentiable and continuous on the interval. If then f’(c) = 1. this means that the derivative must equal 1 and that the values of x are the c values of the function. f’(c) = 1 at c = 2.96 on the interval.

$1,000 On what intervals is the graph of f(x) = -8 / x 3 increasing? B - (0, ∞) A - (-∞, 0) C - f(x) is strictly decreasing D - (-∞, ∞)

Explanation A function is increasing on all the intervals that f’(x) > 0 [ f(x) has a slope that is greater than 0] From the graph of f’(x) you can see that for -inf. 0 and therefore on the interval (-inf., inf.) f(x) is increasing.

$2,000 What is the differential dy? Y = (49 - x 3 ) 1/2 B - dy = (1/2)(-3x 2 )(49 - x 3 ) 1/2 dx A - dy = (1/2)(-3x 2 ) -1/2 dx C - dy = (1/2)(49 - x 3 ) -1/2 dx D - dy = (1/2)(-3x 2 )(49 - x 3 ) -1/2 dx

Explanation To find the differentiable, derive, and then multiply both sides by dx. Y = (49 - x 3 ) 1/2 Deriving you get: dy/dx = (49 - x 3 ) 1/2 Multiply each side by dx to get: dy = (49 - x 3 ) 1/2 dx

$4,000 Determine the points of inflection of the function f(x) = (x 3 +2)(x 4 ) B - (-1.046, 1.024) A - (-0.830, 0.677) C - (1.3 x , 6 x ) D - (-0.760, 0.987) A - (-0.830, 0.677)

Explanation The x values of points of inflection on f(x) exist where f’’(x) = 0. When f(x) = (x 3 +2)(x 4 ) or x 7 +2x 4, f’(x) = 7x 6 + 8x 3. So, f’’(x) = 42x x 2 Using a graphing calculator find where f’’(x) = 0. For 0 = 42x x 2, x = Now, substituting.83 into the original equation, we find that he coordinate of the p of I is (-0.830, 0.677)

$8,000 Find the limit as x app. Inf. f(x) = 20/(x 2 + 1) B - Positive Infinity A - Limit does not exist C - 0 D - Negative Infinity C - 0

Explanation If you divide everything in an equation by x to the highest power in the denominator, then plug in infinity for x, you can find the limit as x approaches infinity. Lim as x app. Infinity 20 / 1+x 2 =

$16,000 On what intervals is the concavity positive on f(x) = -2x 2 (1-x 2 ) B - (-∞, ) (0.408, ∞) A - (-∞, ) (0.707, ∞) C - (-0.408, 0.408) D - (-0.707, 0.707) B - (-∞, ) (0.408, ∞)

Explanation The concavity of f(x) at any value of x is determined by the sign ( + or - ) of f’’(x). If the sign is + then the concavity is positive and negative if the sign is -. Points of infection divide intervals of different concavity. P of I occur where f’’(x) = 0 and f’’(x) = 0 at x = ±0.408 so the intervals of different concavity are (-inf., -.408), (-.408,.408), and (.408, inf.) Explanation cont. >>

Explanation cont. By checking an x value in each interval in the second derivative you find that on (-inf., -.408) and (.408, inf.) the concavity is > 0 And that on (-.408,.408) the concavity < 0

$32,000 The radius of a ball measures 5.25 inches. If the measurement is correct to within 0.01 inch, estimate the propagated error in the volume of the ball. V = (4/3) πr 3 B - ±3.464 in 3 A - ±2.639 in 3 C - ±3.464 in 2 D in 3 B - ±3.464 in 3

Explanation First find the differential dv. dv = 4 π r 2 dr Given is r = 5.25in and dr = ±0.01in so: dv = 4 π (5.25) 2 (±0.01) = ±3.464in 3

$64,000 Find the slope and concavity at x = -7 on y = (-2x 3 )(sin x) B - m = concavity < 0 A - m = concavity > 0 C - m = concavity > 0 D - m = concavity < 0 C - m = concavity > 0 Non-graphing calculator

Explanation The slope at x = -7 can be found by plugging -7 in for x in the derivative of the function. Use the product rule to get f’(x) = (-6x 2 )(sin x) + (-2x 3 )(cos x) f’(-7) = Explanation continued >>

Explanation Cont. Use the sign of f’’(-7) to find concavity f’(x) = (-6x 2 )(sin x) + (-2x 3 )(cos x) f”(x) = (-12x)(sin x) + (-6x 2 )(cos x) + (-6x 2 )(cos x) + (-2x 3 )(-sin x) = Because the second derivative at x = -7 is negative, the concavity at x = -7 is negative.

$125,000 Find the concavity and the equation of the tangent line at x = 3 on y = (2)/(x 2 +x 3 ) B - conc >0 y = x A - conc <0 y = x C - conc >0 y = x D - conc >0 y = x B - conc >0 y = x Non-graphing calculator

Explanation f(x) = (2)/(x 2 +x 3 ) or (2)(x 2 +x 3 ) -1 f(3) =.0556 = y f’(x) = 2(-1(x 2 +x 3 ) -2 (2x +3x 2 )) f’(3) = = m f”(x) = -2((-2(x 2 +x 3 ) -3 )(2x +3x 2 ) + ((x 2 +x 3 ) -2 )(2+6x) f”(3) = f”(3) > 0 so the concavity is positive. Use (y-y 1 )=m(x-x 1 ) >>> (y ) = (-.051)( x - 3 ) Simplify to get y = x

$250,000 A farmer plans to fence a rectangular pasture adjacent to a river. The pasture must contain 180,000m 2 in order to provide enough grass for the herd. When looking for the dimensions that requires the least amount of fencing, what equation should be set to zero to solve for the length of side across from the river? B - y = x +360,000/x A - y = 2x + 180,000/x C - y = ,000/x 2 D - y = ,000/x 2

Explanation Primary: F = 2x + y Secondary: xy = 180,000 We know that y is the side across from the river because there is only one of that side. Therefore we want to be solving for y. x = 180,000 / y. Substitute x into the primary. F = 2(180,000)/y + y Now, \set ,000/y 2 equal to zero to find the minimum value for y.

$500,000 Find the absolute maximum of the function f’’(x) on [-1, 2] f(x) = x 8 + 2x 4 B - ( 2, 3680 ) A - ( 2, 4712 ) C - ( 0, 0 ) D - ( -1, 3890 B - ( 2, 3680 ) Non-graphing calculator

Explanation The absolute maximum of f”(x) is determined by testing all of the critical numbers and endpoints of f”(x). The critical numbers are determined by setting f’’’(x) equal to zero. f(x) = x 8 + 2x 4 f’(x) = 8x 7 + 8x 3 f”(x) = 56x x 2 f’’’(x) = 336x x f’’’(x) = 0 at x = 0 When we plug -1, 0, and 2 into the f(x) we find that 2, 3680 is the absolute maximum on the interval

$1,000,000 The functions f and g are differentiable for all real numbers. The function of h is given by h(x) = f(g(x)) - 6 B - m = -5 A - m = 14 C - m = 5 D - m = -6 B - m = -5 xf(x)f’(x)g(x)g’(x) What slope, h’(r), must exist on 1 < r < 3

Explanation First find h(1) and h(3) and find the slope that is created by the two points. Then, by the definition of the mean value theorem, there must be a point on the interval with that slope. h(x) = f(g(x)) - 6 h’(1) = f(g(1)) - 6 = f(2) - 6 = = 3 h’(3) = f(g(3)) - 6 = f(4) -6 = = -7 m = (y 2 - y 1 ) / (x 2 - x 1 ) = (-7 - 3) / (3 - 1) = -5