Homework_7_Solutions - Math 1110 Name: Homework 7 Due 10/13...

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Unformatted text preview: Math 1110 Name: Homework 7 Due 10/13 or 10/14 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 Text exercises. Please do the following problems from the book. §3.6 #10, 18, 22, 48, 58, 60, 68, 78, 84, 88abf, 98, 104 §3.7 #2, 20, 23, 30, 32, 41, 42, 43, 50 /1 sometimes) f alse, a nd c ircle y our r espons - a r eason w hy i t's t rue, o r a n e xample w (a) I f f (x) i s a n e venf unction,t hen s o i (b) I f f ( x) a nd g (x) b oth o ne-to-one f un Math 1110 (Fall 2011) HW7 Presentation Problems 2 Question 1. (10 points) Determine whether the following statements are true or false, and circle your response. Please give a brief explanation (in a complete sentence!) – a reason why it’s true, or an example where it fails. (a) If g(x) is one-to-one and differentiable at every point in R, then its reciprocal differentiable at every point in its domain. If g(x) is such a function, then let f(x) denote 1 g(x) is also 1 (b) I f f ( x) a nd g (x) b oth o ne-to-one f un . Using the quotient rule, we see that the g(x) g ￿ (x) for all x such that g(x) ￿= 0. This is because g(x) is g(x)2 differentiable at every point in R. The domain of f(x) is all points such that g(x) ￿= 0, and so f(x) is differentiable at every point in its domain. derivative of f exists and is equal to (b) The derivative of an even function is an odd functrion. Let f(x) be an even function. Let a denote a point at which f is differentiable. Then f is differentiable at −a because f is symmetric across the y-axis, and f(−a + h) − f(−a) h→ 0 h f(a − h) − f(a) = lim , because f is even h→ 0 h f(a + k) − f(a) = lim , where k = −h −k→ 0 −k f(a + k) − f(a) = lim k→ 0 −k f(a + k) − f(a) = − lim k→ 0 k ￿ = −f (a) f ￿ (−a) = lim So we conclude that f ￿ (x) is an odd function. A trickier way to solve the question is to notice that f(a) − f(−a) = 0. Differentiating and using the chain rule, we see that f ￿ (a) − (−f ￿ (−a)) = 0. So we conclude that f"(x) is odd. Math 1110 (Fall 2011) HW7 Presentation Problems 3 Question 2. (10 points) Find the equation for the line tangent to the curve 2xy + sin(y) = 2π at the point (1, π). We begin by implicitly differentiating the equation we are given, to get 2xy ￿ + 2y + cos(y)y ￿ = 0 Next we solve for y ￿ when x = 1 and y = π. 0 = 2y ￿ + 2π + cos(π)y ￿ = 2y ￿ + 2π − y ￿ = y ￿ + 2π We conclude that at the point (1, π) we must have that y ￿ = −2π. Now we have the slope of the tangent line we want, and one point on the line, so we solve for the y-intercept. y = −2πx + b π = −2π · 1 + b 3π = b So the equation for the tangent line is y = −2πx + 3π. Homework 7 Book Problem Answers Section 3.6: Section 3.7: ...
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