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NEW 2.3E: Limit Laws

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These are homework exercises to accompany Chapter 2 of OpenStax's "Calculus" Textmap.

3.6 E: Rates of Change Exercises

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In this case, \(s(t)=0\) represents the time at which the back of the car is at the garage door, so \(s(0)=−4\) is the starting position of the car, 4 feet inside the garage. Suppose the price-demand ...In this case, \(s(t)=0\) represents the time at which the back of the car is at the garage door, so \(s(0)=−4\) is the starting position of the car, 4 feet inside the garage. Suppose the price-demand and cost functions for the production of cordless drills is given respectively by \(p=143−0.03x\) and \(C(x)=75,000+65x\), where \(x\) is the number of cordless drills that are sold at a price of \(p\) dollars per drill and \(C(x)\) is the cost of producing \(x\) cordless drills.

3.2E: Derivative as a Function Exercises

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102) [T] The best linear fit to the data is given by \(H(t)=7.229t−4.905\), where \(H\) is the height of the rocket (in meters) and t is the time elapsed since takeoff. 104) [T] The best cubic fit to ...102) [T] The best linear fit to the data is given by \(H(t)=7.229t−4.905\), where \(H\) is the height of the rocket (in meters) and t is the time elapsed since takeoff. 104) [T] The best cubic fit to the data is given by \(F(t)=0.2037t^3+2.956t^2−2.705t+0.4683\), where \(F\) is the height of the rocket (in m) and \(t\) is the time elapsed since take off.

4.3E: Shape of the Graph Exercises

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Decreasing for \(−2<x<−1\) and \(1<x<2\); increasing for \(−1<x<1\) and \(x<−2\) and \(x>2\) maxima at \(x=−1\) and \(x=1\), minima at \(x=−2\) and \(x=0\) and \(x=2\) 216) \(f(x)>0,f′(x)>0\) over \(x...Decreasing for \(−2<x<−1\) and \(1<x<2\); increasing for \(−1<x<1\) and \(x<−2\) and \(x>2\) maxima at \(x=−1\) and \(x=1\), minima at \(x=−2\) and \(x=0\) and \(x=2\) 216) \(f(x)>0,f′(x)>0\) over \(x>1,−3<x<0,f′(x)=0\) over \(0<x<1\) 217) \(f′(x)>0\) over \(x>2,−3<x<−1,f′(x)<0\) over \(−1<x<2,f''(x)<0\) for all \(x\) 218) \(f''(x)<0\) over \(−1<x<1,f''(x)>0,−3<x<−1,1<x<3,\) local maximum at \(x=0,\) local minima at \(x=±2\)

4.2E: Maxima and Minima Exercises

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Since the absolute maximum is the function (output) value rather than the x value, the answer is no; answers will vary 106) Absolute maximum at \(x=4,\) absolute minimum at \(x=−1,\) local maximum at ...Since the absolute maximum is the function (output) value rather than the x value, the answer is no; answers will vary 106) Absolute maximum at \(x=4,\) absolute minimum at \(x=−1,\) local maximum at \(x=−2,\) and a critical point that is not a maximum or minimum at \(x=2\) 107) Absolute maxima at \(x=2\) and \(x=−3\), local minimum at \(x=1\), and absolute minimum at \(x=4\) Absolute minima: \(x=0, x=2, y=1\); local maximum at \(x=1, y=2\)

5.1 Approximating Area (Riemann Sum) Exercises

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But, \(\displaystyle f(c)≤f(x)\) for \(\displaystyle c≤x≤d\), so the area under the graph of \(\displaystyle f\) between \(\displaystyle c\) and \(\displaystyle d\) is \(\displaystyle f(c)(d−c)\) plus...But, \(\displaystyle f(c)≤f(x)\) for \(\displaystyle c≤x≤d\), so the area under the graph of \(\displaystyle f\) between \(\displaystyle c\) and \(\displaystyle d\) is \(\displaystyle f(c)(d−c)\) plus the area below the graph of f but above the horizontal line segment at height \(\displaystyle f(c)\), which is positive.

4.E: Open Stax 4.1 - 4.5 Exercises

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These are homework exercises to accompany Chapter 4 of OpenStax's "Calculus" Textmap.

2.5E: Limits at Infinity EXERCISES

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For the following functions \(f(x)\), determine whether there is an asymptote at \(x=a\). 263) \(\displaystyle \lim_{x→∞}\frac{x^2−2x+5}{x+2}\) 264) \(\displaystyle \lim_{x→−∞}\frac{3x^3−2x}{x^2+2x+8}...For the following functions \(f(x)\), determine whether there is an asymptote at \(x=a\). 263) \(\displaystyle \lim_{x→∞}\frac{x^2−2x+5}{x+2}\) 264) \(\displaystyle \lim_{x→−∞}\frac{3x^3−2x}{x^2+2x+8}\) 267) \(\displaystyle \lim_{x→−∞}\frac{\sqrt{4x^2−1}}{x+2}\) 270) \(\displaystyle \lim_{x→∞}\frac{2\sqrt{x}}{x−\sqrt{x}+1}\) 271) \(f(x)=x−\frac{9}{x}\) 279) \(f(x)=\frac{1}{x^3+x^2}\) Horizontal: \(y=0,\) vertical: \(x=0\) and \(x=−1\) CR 2) \(\displaystyle \lim_{x→∞}cos(\frac{1}{x})\)

2.3E: Limit Laws and Techniques for Computing Limits EXERCISES

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\(\displaystyle \lim_{x→4}\frac{x^2−16}{x−4}=\frac{16−16}{4−4}=\frac{0}{0}; then, \displaystyle \lim_{x→4}\frac{x^2−16}{x−4}= \displaystyle \lim_{x→4}\frac{(x+4)(x−4)}{x−4}=8\) \( \displaystyle \lim _...\(\displaystyle \lim_{x→4}\frac{x^2−16}{x−4}=\frac{16−16}{4−4}=\frac{0}{0}; then, \displaystyle \lim_{x→4}\frac{x^2−16}{x−4}= \displaystyle \lim_{x→4}\frac{(x+4)(x−4)}{x−4}=8\) \( \displaystyle \lim _{x→−9}(xf(x)+2g(x))=( \displaystyle \lim _{x→−9}x)( \displaystyle \lim _{x→−9}f(x))+2 \displaystyle \lim _{x→−9}(g(x))=(−9)(6)+2(4)=−46\)

MTH 210 Calculus I

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4.0E: Exercises

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4.0E: Exercises 0) Did you read the Chapter 4 Prelude, section 4.0? Answer: Cool Rocket

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