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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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