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1) y = x2 + 4x, [4, 7] A) 11 B) 45 7 C) 77 3 D) 15 1) 2) y = 2x, [2, 8] A) - 3 10 B) 1 3 C) 7 D) 2 2) Use the slopes of UQ, UR, US, and UT to estimate the rate of change of y at the specified value of x. 3) x = 5 1 2 3 4 5 6 x y 5 4 3 2 1 Q R S T U 1 2 3 4 5 6 x y 5 4 3 2 1 Q R S T U A) 1 B) 2 C) 5 D) 0 3) Use the table to estimate the rate of change of y at the specified value of x. 4) x = 1. x y 0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 0 0.02 0.08 0.18 0.32 0.5 0.72 0.98 A) 2 B) 0.5 C) 1 D) 1.5 4) Use the graph to evaluate the limit. 5) lim x→-1 f(x) -6 -5 -4 -3 -2 -1 1 2 3 4 5 6 x y 1 -1 A) 3 4 B) ∞ C) -1 D) - 3 4 5) Find the limit. 6) lim x→2 (8x + 3) A) 3 B) -13 C) 11 D) 19 6) 1) y = x3 2 , (8, 256) A) y = 32x - 512 B) y = 32x + 512 C) y = 512x + 96 D) y = 96x - 512 1) Graph the equation and its tangent. 2) Graph y = 5x2 and the tangent to the curve at the point whose x-coordinate is 1. -5 5 x y 10 -10 -5 5 x y 10 -10 A) -5 5 x y 10 -10 -5 5 x y 10 -10 B) -5 5 x y 10 -10 -5 5 x y 10 -10 C) -5 5 x y 10 -10 -5 5 x y 10 -10 D) -5 5 x y 10 -10 -5 5 x y 10 -10 2) Calculate the derivative of the function. Then find the value of the derivative as specified. 3) f(x) = 5x + 9; f ′ (2) A) f ′ (x) = 5; f ′ (2) = 5 B) f ′ (x) = 0; f ′ (2) = 0 C) f ′ (x) = 9; f ′ (2) = 9 D) f ′ (x) = 5x; f ′ (2) = 10 3) 4) g(x) = 3x2 - 4x; g ′ (3) A) g ′ (x) = 6x; g ′ (3) = 18 B) g ′ (x) = 3x - 4; g ′ (3) = 5 C) g ′ (x) = 2x- 4; g ′ (3) = 2 D) g ′ (x) = 6x - 4; g ′ (3) = 14 4) Find the indicated derivative. 5) dy dx if y = 3x3 A) 9x B) 3x2 C) 9x2 D) 9x3 5) 1) y = 14 - 12x2 A) 14 - 12x B) 14 - 24x C) -24x D) -24 1) 2) y = 5x4 + 7x3 - 4 A) 20x3 + 21x2 B) 4x3 + 3x2 - 7 C) 4x3 + 3x2 D) 20x3 + 21x2 - 7 2) Find the second derivative. 3) y = 9x2 + 3x - 7 A) 18x + 3 B) 9 C) 18 D) 0 3) 4) y = 6x4 - 7x2 + 2 A) 24x2 - 14x B) 72x2 - 14x C) 72x2 - 14 D) 24x2 - 14 4) Find y ′ . 5) y = (5x - 5)(6x + 1) A) 60x - 12.5 B) 60x - 35 C) 30x - 25 D) 60x - 25 5) Find the derivative of the function. 6) y = x2 - 3x + 2 x7 - 2 A) y ′ = -5x8 + 18x7 - 14x6 - 4x + 6 (x7 - 2)2 B) y ′ = -5x8 + 18x7 - 14x6 - 3x + 6 (x7 - 2)2 C) y ′ = -5x8 + 18x7 - 13x6 - 4x + 6 (x7 - 2)2 D) y ′ = -5x8 + 19x7 - 14x6 - 4x + 6 (x7 - 2)2 6) 1) y = x2 + 4x, [4, 7] A) 11 B) 15 C) 45 7 D) 77 3 1) Find the slope of the curve at the given point P and an equation of the tangent line at P. 2) y = x2 + 5x, P(4, 36) A) slope is - 4 25 ; y = - 4x 25 + 8 5 B) slope is 13; y = 13x - 16 C) slope is 1 20 ; y = x 20 + 1 5 D) slope is -39; y = -39x - 80 2) Use the slopes of UQ, UR, US, and UT to estimate the rate of change of y at the specified value of x. 3) x = 5 1 2 3 4 5 6 x y 5 4 3 2 1 Q R S T U 1 2 3 4 5 6 x y 5 4 3 2 1 Q R S T U A) 5 B) 0 C) 1 D) 2 3) Use the table to estimate the rate of change of y at the specified value of x. 4) x = 1. x y 0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 0 0.02 0.08 0.18 0.32 0.5 0.72 0.98 A) 1 B) 1.5 C) 0.5 D) 2 4) 1 Use the graph to evaluate the limit. 5) lim x→-1 f(x) -6 -5 -4 -3 -2 -1 1 2 3 4 5 6 x y 1 -1 A) - 3 4 B) -1 C) 3 4 D) ∞ 5) Find the limit. 6) lim x→2 (8x + 3) A) 19 B) 3 C) -13 D) 11 6) Find the limit if it exists. 7) lim x→-3 (4x - 4) A) -16 B) 16 C) 8 D) -8 7) Find the limit. 8) lim x→0 (3 sin x - 1) A) 3 B) 0 C) -1 D) 3 - 1 8) Find an equation for the tangent to the curve at the given point. 9) y = x3 2 , (8, 256) A) y = 32x - 512 B) y = 96x - 512 C) y = 512x + 96 D) y = 32x + 512 9) Calculate the derivative of the function. Then find the value of the derivative as specified. 10) g(x) = 3x2 - 4x; g ′ (3) A) g ′ (x) = 6x; g ′ (3) = 18 B) g ′ (x) = 3x - 4; g ′ (3) = 5 C) g ′ (x) = 2x- 4; g ′ (3) = 2 D) g ′ (x) = 6x - 4; g ′ (3) = 14 10) Find the indicated derivative. 11) dy dx if y = 3x3 A) 9x3 B) 3x2 C) 9x2 D) 9x 11) Find the derivative. 12) y = 2 - 8x3 A) -24x B) 2 - 24x2 C) -16x2 D) -24x2 12) Find y ′ . 13) y = (5x - 5)(6x + 1) A) 60x - 12.5 B) 60x - 25 C) 30x - 25 D) 60x - 35 13) Find the derivative of the function. 14) y = x3 x - 1 A) y ′ = -2x3 - 3x2 (x - 1)2 B) y ′ = 2x3 + 3x2 (x - 1)2 C) y ′ = -2x3 + 3x2 (x - 1)2 D) y ′ = 2x3 - 3x2 (x - 1)2 14) Find the derivative. 15) y = 8 x + 3 sec x A) y ′ = - 8 x2 + 3 sec x tan x B) y ′ = - 8 x2 - 3 csc x C) y ′ = 8 x2 - 3 sec x tan x D) y ′ = - 8 x2 + 3 tan2x 15)

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Problems In Calculus And Their Solutions
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