Homework_5_Solutions - Math 1110 Name: Homework 5 Due 9/29...

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Unformatted text preview: Math 1110 Name: Homework 5 Due 9/29 or 9/30 in class Please print out these pages. Write your answers to the “text exercises” on separate paper and staple it to these pages. You should include computational details. These problems will be assessed for completeness. Always write neatly and legibly. Please answer the “presentation problems” in the spaces provided. Include full explanations and write your answers in complete, mathematically and grammatically correct sentences. Your answers will be assessed for style and accuracy, and you will be given written feedback on these problems. GRADES Text exercises / 20 Pres probs / 20 Staple /1 Text exercises. Please do the following problems from the book. §3.1 #8, 22, 23, 30, 34, 36 §3.2 #2, 8, 16, 27, 28, 29, 30, 36, 42, 46 Question 1. (10 points) Does knowing that a function g(t) is differentiable at t = 50 tell you anything about the differentiability of the function −2g(t) at t = 50? Give reasons for your answer! g(50 + h) − g(50) Since g(t) is differentiable at t = 50 we know that lim exists. We want to h→ 50 h −2g(50 + h) − (−2)g(50) consider lim . If this limit exists then −2g(t) is differentiable at t = 50. h→ 50 h Note: −2g(50 + h) − (−2)g(50) g(50 + h) − g(50) = lim −2 h→ 50 h→ 50 h h −2g(50 + h) − (−2)g(50) = −2 lim h→ 50 h lim −2g(50 + h) − (−2)g(50) = g ￿ (50), and we know h that g is differentiable at t = 50. So we conclude that −2g(t) is differentiable at t = 50, and that its derivative at that point is −2g ￿ (50). We know that the final limit exists because lim h→ 50 Math 1110 (Fall 2011) HW5 Presentation Problems 2 Question 2. (10 points) Find all points on the curve y = 2x3 + 3x2 − 12x + 1 where the tangent line is horizontal. Give equations for all such tangent lines. We want to find points on the given curve where the derivative dy = 0, and then find the tangent dx lines to the curve at these points. We begin by computing dy/dx. dy 2(x + h)3 + 3(x + h)2 − 12(x + h) + 1 − (2x3 + 3x2 − 12x + 1) = lim dx h→ 0 h 3 + 3x2h + 3xh2 + h3) + 3(x2 + 2xh + h2) − 12(x + h) + 1 − (2x3 + 3x2 − 12x + 1) 2(x = lim h→ 0 h 2h + 3xh2 + h3) + 3(2xh + h2) − 12h 2(3x = lim h→ 0 h = lim (2(3x2 + 3xh + h2) + 3(2x + h) − 12) h→ 0 2 = 6x + 6x − 12 Now we plug in dy dx = 0 and solve for x: 0 = 6x2 + 6x − 12 = 6(x2 + x − 2) = 6(x − 1)(x + 2) So in order for dy to equal 0, we must have x = 1 or x = −2. Plugging x = 1 into the equation for dx our curve, we see that the only y-value corresponding to x = 1 is 2(1)3 + 3(1)2 − 12(1) + 1 = −6. Similarly, if x = −2, then y = 21. We want to find the equation of a horizontal line going through the point (1, −6) and an equation for a horizontal line that goes through the point (2, 5). The only such equations are y = −6 and y = 21, respectively. Homework 5 Book Problem Answers Section 3.1: Section 3.2: ...
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