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Math 141 Exam #2 Study Guide

by: ErinNordquist

Math 141 Exam #2 Study Guide 141

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About this Document

This study guide covers everything we learned in class in sections 7.1, 7.2, 7.3, 7.5, 7.6, and 8.1 which is what will be on the test this Friday.
Calculus II
Wisely Wong
Study Guide
Math, Calulus, calc2, Study Guide
50 ?




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This 6 page Study Guide was uploaded by ErinNordquist on Sunday October 2, 2016. The Study Guide belongs to 141 at University of Maryland taught by Wisely Wong in Fall 2016. Since its upload, it has received 96 views. For similar materials see Calculus II in Math at University of Maryland.


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Date Created: 10/02/16
Math141 Sections:7.1,7.2,7.3,7.5,7.6,8.1 ExamDate:Friday,October7,2016 Section 7.1: Inverse Functions ● DEFINITION​: If f is a “one-to-one” function, then f has an inverse provided there is a function g such that the domain of g is the range of f such that f(x) = y if and only if g(y) = x for all x in the domain of f and all y in the range of f. If this is true, g is the −1 inverse of f and is designated by f . ● DEFINITION​: Horizontal Line Test - a function f has an inverse if and only if for any two numbers in the domain: if x = / x then f(x ) / f(x ) (if you can draw a horizontal 1 2 1 2 line that touches the line twice, it fails the test) ○ Any function that is strictly increasing or decreasing has an inverse ● Simplified definition: For the function to have an inverse, no two x-values can yield the same y-value(vertical line test). If this isn’t true over the entire function, you must restrict the domain to a segment where it is true. The inverse of f is the function that will give you the x-value of the point (x,y) when y is put in, compared to the original which will give you v when x is put in. −1 −1 ○ f(x) = y and f (y) = x but the inverse is usually denoted as f (x) = y . For the inverse, the “x” refers to the y-coordinate and the “y” refers to the x-coordinate that you’d get from the original function. ○ Pictographically: ○ 1 Finding the Inverse ● Process: 1. y = f(x) 2. Solve for x in terms of y −1 3. Rewrite as f (y) = x −1 4. Switch x and y to get f (x) = y ● Note: you can also switch them at the beginning and solve for y Inverse Properties −1 −1 −1 ● f has an inverse and (f ) = f −1 ● f (f(x)) = x for all x in domain of f ● f(f (y)) = y for all y in the range of f ● Suppose that f has an inverse and is continuous on an open interval I containing a. Assume that f ′(a) exists, does not equal 0, and f(a) = c. Then (f )−1′(c) exists and −1 1 ○ FORMULA: (f )′(c) = f′(a) ● Given f(x) and its inverse graphed on the same axes, they will be reflections on each other about the line y = x. ● The domain of the original function, where it is invertible, is the range of the inverse. The range of the original function is the domain of the inverse. **Practice Problems on pg. 439 (suggested #5, 18, 28, 35, and 41) Section 7.2: The Natural Exponential Function Definition of Work x ● DEFINITION: ​ Natural logarithmic function: ln(x) = 1dt so d ln(x) = (x) 1t dx x ● DEFINITION​: Natural exponential function: e is the inverse of the natural log function ln(x). So y = e ⇔ x = ln(y) Properties 2 ln(x) x ● e = x = ln(e ) ● ln(x)had domain (0,∞) and range (− ∞,∞) ● e has domain (− ∞,∞) and range (0,∞) **Practice Problems on pg. 446 (suggested #3, 5, 8, 11, 14, 17, 20, 25, and 28) Section 7.3: ​eneral Exponential and Logarithmic Functions Exponential Functions ● DEFINITION​: We consider exponential function a where “a ” is the base, and “x” is the exponent. ○ a = [(eln(a) ] = e ln(a) ● For the derivative, d x d x ln(a) x ln(a) ○ dx(a ) =dx(e ) = e * ln(a) x x ln(a) d x x ○ a = e so ​ ORMULA​: dxa = a * ln(a) ● For the Integral d x x x a ○ dxa = a * ln(a) so ​ ORMULA​: a =∫ ln(a) C Logarithmic Functions ● DEFINITION:​ let a > 0, a = / 1. The logarithm, base “a” is given by log (x).(Read as “log a base a of x”) ○ Note: log (e) = ln(x) y ○ y = loga(x) ⇔ x = a ○ ● Properties ln(x) ○ logax = ln(a) ○ For any x, y > 0; log (ay) = log ax) + log ay) 3 ○ For any real number “r”; r log (xa = log (xa) r x −1 −1) ○ So, by the last two: log (a)y= log (ay ) = log (xa + log (a = loga(x) − loa (y) **Practice Problems on pg. 454 (suggested #1, 4, 17, 18, 33, and 34) Section 7.5: Inverse Trigonometric Functions Sine ● DEFINITION: ​ the inverse sine function, denoted sin (x) or arcsin(x) is −1 −π π ○ y = sin (x) ⇔ sin(y) = x, while 2 ≤ y ≤ 2 where y is the input, and − 1 ≤ x ≤ 1 where x is the output. ○ (This is the typical domain restriction of sin to make sure it passes the Horizontal Line test.) Cosine −1 ● DEFINITION​: the inverse cosine function, denoted cos (x) or arccos(x) is ○ y = cos (x) ⇔ cos(y) = x, while 0 ≤ y ≤ π where y is the input, and − 1 ≤ x ≤ 1 where x is the output. ○ (This is the typical domain restriction of cos to make sure it passes the Horizontal Line test.) Derivatives and Integrals ● dxsin (x)) = √ 1 2 and dx(cos (x)) = √−12 can be derived using implicit differentiation. 1−x 1−x ● Using those equations we get… ○ FORMULA:​ ∫ 1 dx = sin ( ) + C --typically used-- √a −x2 a ○ FORMULA​: ∫ 1 dx =− cos ( ) + C √a −x2 a Tangent −1 ● DEFINITION​: the inverse tangent function, denoted tan (x) or arctan(x) is ○ y = tan (x) ⇔ tan(y) = x when − ∞ < x < ∞ and 2π < y <π2 4 ● To find the derivative: ○ Rewrite tan (x) = y as tan(y) = x, then use implicit differentiation… ○ sec (y* dy = 1 ⇒ dy= 2 = 12 dx dx sec (y)1+tan (y) ○ Becuase tan(y) = x… ○ dytan (x) = 1 which can be generalized as dx 1+x ○ FORMULA​: ∫ 2 2dx = aan ( a + C √x +a **Practice Problems on pg. 470 (suggested #1-9, 14-17, 19-24, 28, 30, 35, 38, 39, and 42) Section 7.6: L’Hôpital’s Rule Mean Value Theorem Review f(x) ● DEFINITION:​ we consider limit lix→ag(x) where BOTH f(x) and g(x) are approaching 0 or both are approaching ± ∞ creating 0 or ±∞ . These are known as ​indeterminate forms. ● DEFINITION:​ Let f(x) and g(x) be differentiable “around” x = a, and assure g (x) =/ 0 “around” x = a. If… ○ {lim f(x) = 0 AND limg(x) = 0} - OR - {limf(x) =± ∞ AND limg(x) =± ∞} x→a x→a x→a x→a f(x) f′x ) ○ Then…​FORMULA:​ lim = lim x → a g(x) x→ a g′(x) ● Note: this may not be the most efficient method in all cases **Practice Problems on pg. 481 (suggested #2, 11, 12, 16, 25, 27, 42 43, 46, and 49) Section 8.1: Integration by Parts derived from the product rule… ● The product rule: d (f(x)g(x)) =(x)g(x) + f(x′(x) dx 5 ● So… f(x)g(x) = f∫′(x)g(x) dx ∫ f(x′(x) dx ● Rewritten as...∫f(x)g(x) dx = (f(x)g(x))∫′(x)g(x) dx...then with usual notation ● FORMULA​: ∫u dv = (u v)*− v du∫ ○ Note: the function that is easier to integrate should be used as g ′(x)or "dv" ● This formula can also be used for definite integrals. b b ○ ∫u dv = (u v)| − v du ∫ a * a a ● Integration by parts can be used multiple times on the same integral to reduce powers (especially of Trig functions using these identities) 1−cos(2x) ○ sin (x) = 2 ○ cos (x) = 1+cos(2x) 2 **Practice Problems on pg. 509 (suggested #1, 5, 6, 8, 9, and 12) Other Useful Information ● Trigonometric Identities ○ sin2x ) +cos 2x ) = 1 ○ sin (x) = 1−cos(2x) 2 2 1+cos(2x) ○ cos (x) = 2 ○ sin(2x) = 2sinx cos (x) ○ cos(2x) = (cos2(x) −sin2(x)) = (cos 2x ) + 1) = (1 −sin2(x)) ● Remember: ○ Make sure you are aware of when you need to take the derivative vs the integral. ○ Before using L’Hopital’s Rule: always make sure f(x) and g(x) are approaching the same thing, otherwise you may need to rewrite the integral so they are. 6


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