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MATH 153 Exam #2 Study Guide

by: Sophia Valla

MATH 153 Exam #2 Study Guide MATH 153

Marketplace > West Virginia University > Math > MATH 153 > MATH 153 Exam 2 Study Guide
Sophia Valla
GPA 3.5

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Things to expect on the next exam.
Calculus 1A with Precalculus
Andy Blankenship
Study Guide
Math, Calculus, Precalculus
50 ?




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This 6 page Study Guide was uploaded by Sophia Valla on Sunday September 25, 2016. The Study Guide belongs to MATH 153 at West Virginia University taught by Andy Blankenship in Fall 2016. Since its upload, it has received 15 views. For similar materials see Calculus 1A with Precalculus in Math at West Virginia University.


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Date Created: 09/25/16
Math 153 Exam #2 Study Guide Average Rate of Change or Average Velocity ● Slope avg velocity y❑ 2y❑ 1→ f (b)− f (→(a, f (a))/(b, f (b)) ○ x❑ 2x❑ 1 b−a Limits ● Left Hand Limits −¿ x→a❑ f (x) ○ lim ¿ ¿ ● Right Hand Limits x→a❑ f (x) ○ lim ¿ ¿        x→a+¿ ? lim f (x)=L if and only if x→a−¿ f (x)=L=lim ¿ x→a lim ¿ ¿ End Behavior ● Horizonatl asymptotes ○ y = L is a H.A.  ○ x→ ∞f (x)=L  or  x→−∞ f (x)=L ● Vertical asymptotes ○ x = a is a V.A. x→a−¿ f (x)=+¿−∞ ○ lim F(x)=+¿−∞ or or x→a lim ¿ ¿ x→a+¿ f (x)=+¿−∞ lim ¿   ¿ Limits Algebraically  ● If  lim f (x),lim g(x) exists and C is a constant  x→a x→ a ○ lim [ f (x)+¿−g(x)]=lim f (x)+¿−lim g(x) x→a x→a x→a lim cf (x)=c∗lim f (x) ○ x→a x→ a lim [ f (x)g(x)]=lim f (x)∗lim g(x) ○ x→a x→a x→ a lim f (x) f (x) x→a ○ lim [g(x) = lim g(x)   ;(denominator  ≠0 ¿ x→a x→a ? If f is a polynomial or rational function and a is in the domain of f then  lim f (x)= f (a) x→a 0 Limits 2nd Case “ ” Form 0 0 ● Graphically,  lim f (x) yields  (hole in the graph) x→a 0 x→a+¿ f (x) ● x→a−¿ f (x)=lim¿¿ only if  lim f (x)≠ f (a)  the limit still exists lim ¿ x →a ¿ A. Factoring / Cancelling ­  Solution by direct substitution of a (if sol→ ido more steps to find limit) ­ Simplify by factoring  ­ Plug a into factored expression (gives you limit)           B. Using Conjugates for Square Roots ­ Direct substitution of a (if soln is 0→0  do more steps to find limit) ­ Simplify by multipplying by the conjugate  ­ Foil out expression ­ Simplify numerator ­ denominators remains the same ­ Simply numerator by factoring  ­ Plug a into factored expression (gives you limit)            C. Common Denominators ­ Solution by direct substitution of a (if soln is 0→0  do more steps to find limit) ­ Find a common denominator ­ Multiply top and bottom to have the same common denominator ­ Simply numerator by factoring  ­ Plug a into factored expression (gives you limit) Precise Definition of the Limit ●  “For every positive Epsilon ( ε > 0) there is a positive deltaδ>0¿ such that if               0 < | <  δ , then  | f (x)−L| <  ε  “ lim f (x)=L ● x→a Continuity ● A function is continuous at x = a if   lim f (x)= f (a) x→a ● If it is not continuous at a, the f is discontinuous at a ( or has a discontinuity at a) 1. Hole / Removable Discontinuity  lim f (x) lim f (x)≠ f (a) ­ When  x→a exists , but x→a ­ f(a) may or may not be defined  2. Jump / Non Removable x→a+¿ f (x) x→a−¿ f (x) ­ When  lim ¿ ≠lim ¿ ¿ ¿ ­ lim f (x) DOES NOT EXIST x→a 3. Infinite Discontinuity / Nonremovable Discontinuity  x→a−¿ f (x)=+¿−∞ x→a+¿ f (x)=+¿−∞ ­ When  lim ¿ and  lim ¿ ¿ ¿ x→a−¿ f (x)= f (a) ­ F is cont. From the LEFT at a if  lim ¿ ¿ x→a+¿ f (x)= f (a) ­ Fis cont. From the RIGHT at a if  lim ¿ ¿ ­ If f(x)  and g(x) are cont at x=a then so are: ­ (f+g)(x) ­ (f­g)(x) ­ c*f(x) ­ (fg)(x) ­ (f/g)(x) ­ If f is cont at a, g is cont at f(a), then∘g f(x) is cont at a ? All conts for every element in their domain polynomials, rationals, radicals,  trigonometric, exponential, logarithmic, inverse trig.   Intermediate Value Theorem  ● If f is a cont function on the closed interval [a,b] and there is a number N between f(a) and f(b), where f(a)  ≠ f(b) there must be a c between a and b Limits with Infinities (V.A.) ● Can’t factor/cancel ● Can't conjugate/cancel ● Can't find a common denominator/cancel ¿ ­ Direct substitution to get  0 which is undefined (means the limit is usually ∞ or ­ ∞ ) ● If the numerator stays fixed and the deominator shrinks: ○ 5/1=5 5/0.5 = 10  5/0.25= 20  0−¿=−∞ 3x 6 x→2−¿ = ¿ x−2 lim ¿ ¿ 0+¿=∞ 3 x 6 x →2+¿ = x−2 ¿ lim ¿ ¿ ? Both make a V.A. at x = 2 End Behavior (limits as →∞ ) ● If the expression is a fraction, divide every term by x of highest degree in the  denominator ● Gives H.A. as the limit End Behavior of Other Functions lim e❑ =∞ lim e❑ =0 ●   x→ ∞         x→−∞  (H.A. @ y=0) lim e❑ =0 ❑ lim e❑ =∞ ● x→ ∞  (H.A. @ y=0)     ❑  x→−∞ x→0+¿ln x=−∞ ● lim ln x=∞          (V.A. @ x=0) x→ ∞ lim ¿ ¿


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