If f x 3x 2 2x

• THE DIFFERENCE QUOTIENT

  −f(x)becomes−(x2+2x)whichbecomes−x2−2x Now the top floor becomes the following: f(x+h)−f(x)= x2+2xh+h2+2x+2h−x2−2x Simplifying: x2+2xh+h2+2x+2h−x2−2x 2 2 2 222 22 xxhh2xhx2 hhh x x − = +− + +++ + Put this result back on the top floor over the “h” room on the bottom floor. 2 2 ()()22 Factor the common factor of "h" in ...

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• [PDF File]22 Rings of Polynomials - UCI Mathematics

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  A degree n polynomial f(x) 2R[x] is monic if an = 1 (requires R to have a unity). The set of all such polynomials is denoted R[x], the ring of polynomials with coefficients in R. Examples f(x) = 3x2 +2x +1 is a degree two polynomial in the ring Z 4[x]. g(x) = px3 +(1 +i)x is a degree three polynomial in the ring C[x].

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• [PDF File]Rules for Finding Derivatives - Whitman College

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  3.1 The Power Rule 57 Now without much trouble we can verify the formula for negative integers. First let’s look at an example: EXAMPLE3.1.2 Find the derivative of y= x−3.Using the formula, y′ = −3x−3−1 = −3x−4. Here is the general computation.

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• [PDF File]Name: Date: =2 =1−4

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  7. f( x + 1) 8. g(3a) 9. h( x – 2) For 10-11: Let f(x) = -3x + 7 and g(x) = 2x2 – 8. Compute the following: 10. f(g(x)) 11. (g f)(x) 12. find 3 If ( ) 3 5 and ( ) 2, f g f x x g x x $ 13. find 10 If and( ) 9 , f g f gx $

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• [PDF File]The Algebra of Functions

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  The Algebra of Functions . Like terms, functions may be combined by addition, subtraction, multiplication or division. Example 1. Given f ( x ) = 2x + 1 and g ( x ) = x2 + 2x – 1 find ( f + g ) ( x ) and ( f + g ) ( 2 )

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• [PDF File]Limits using L’Hopital’s Rule (Sect. 7.5) 0 L’Hˆopital’s ...

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  x→∞ (3x)2/x. Solution: This limits is of the form ∞0. So, before using L’Hopital’s rule we need to rewrite the function above. (3x)2/x = eln (3x)2/x = e 2 x ln(3x). Since exp is a continuous function, holds lim x→∞ (3x)2/ x= e limx→∞ 2 x ln(3 ) = e x→∞ 2ln(3x) x. The exponent, is an indeterminate limit ∞ ∞. L’Hopital ...

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• [PDF File]Functions - Algebra of Functions

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  3x2 +3 OurSolution Example 6. f(x)=2x − 1 g(x)= x +4 Writeasaproductoffunctions Find(f · g)(3x) f(3x)g(3x) Replacex in f(x) and g(x) with3x [2(3x) − 1][(3x)+4] Multiplyour2(3x) (6x − 1)(3x +4) FOIL 18x2 + 24x − 3x − 4 Combineliketerms 18x2 + 21x − 4 OurSolution

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• [PDF File]Discrete MathematicsDiscrete Mathematics CS 2610

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  Show that f(x) = x2 + 2x + 1 is O(x2). When x > 1 we know that x ≤x2 and 1 ≤x2 then 0 ≤x2 + 2x + 1 ≤x2 + 2x2 + x2 = 4x2 so, let C = 4 and k = 1 as witnesses, i.e., f(x) = x2 + 2x + 1 < 4x2 when x > 1 Could try x > 2. Then we have 2x ≤x2 & 1 ≤x2 then 0 then 0 ≤x2+ 2x + 1 ≤x2+ x2+ x2= 3x2 so, C = 3 and k = 2 are also witnesses to ...

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• [PDF File]Section 2.1: Vertical and Horizontal Asymptotes

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  Example 3. Find the vertical and horizontal asymptotes of the graph of f(x) = x2 2x+ 2 x 1. Solution. The vertical asymptotes will occur at those values of x for which the denominator is equal to zero: x 1 = 0 x = 1 Thus, the graph will have a vertical asymptote at x = 1. To nd the horizontal asymptote, we note that the degree of the numerator ...

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• [PDF File]Composition Functions - University of New Mexico

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  f(x) = 3x2 + 2x+ 1 (2) What do we end up doing with this function? All we do is plug in various values of x into the function because that’s what the function accepts as inputs. So we would have di erent outputs for each input: f( 2) = 3( 2)2 + 2( 2) + 1 = 12 4 + 1 = 9 (3)

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• [PDF File]Chapter 10 Differential Equations

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  2 x2 +C= (i) y= 3x2 + 1 2 (ii) y= 3 2 x2 + 1 2 (iii) y= 5 2 x2 + 1 2 (d) Particular Solution, Different Notation. Solve f′(x) = 3x, given f(1) = 2. Since f′(x) = 3xis just dy dx = 3x, particular solution is still (i) f(x) = 3x2 + 1 2 (ii) f(x) = 3 2 x2 + 1 2 (iii) f(x) = 5 2 x2 + 1 2 (e) Another Particular Solution. Solve dy dx = 3xfor yat ...

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• [PDF File]Answer Key - East Brunswick Public Schools

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  33. Let f be a function that is everywhere differentiable. The value of f' (x) is given for selected values of x in the table below. -10 -5 0 5 10 If f '(x) is always increasing, which statement about f(x) must be true? (A) f(x) has a relative minimum at x = 0. f(x) is concave downwards for all x. has a point of inflection at (0, f (0)).

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• [PDF File]Integration - Past Edexcel Exam Questions

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  19. 2x6 x3 + 3x43 + c 20. f(x) = 4x3 4x2 + x+ 9 21. 2x4 + 4x32 5x+ c 22. (a) y = 3 2 x2 10x12 2x+ 9 (b) (Di erentiation Question) 23. y = 10x12 + 2 5 x5 2 + 11 5 24. (a) (Di erentiation Question) (b) 1 2 x4 13x + c 25. 2x6 2x4 + 3x+ c 26. f(x) = x3 2x32 7x+ 2 27. 2x+ 5 3 x3 + c 28. (a) (Di erentiation Question) (b) y = 1 3 x3 + 6x 9x 4 29. x3 ...

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• [PDF File]Graph Transformations

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  4.) f(x 2) D.) x8 + 2 5.) 1 3 f(x) E.) (x 3) 8 6.) f(3x) F.) x8 7.) f(x) 2 G.) (x 2)8 8.) f(x) H.) (3x)8 9.) f(x + 2) I.) 3x8 10.) f(x 3) J.) (x + 2)8 For #11 and #12, suppose g(x) = 1 x. Match each of the numbered functions on the left with the lettered function on the right that it equals. 11.) 4g(3x 7) + 2 . A.) 6 2x+5 3 12.) 6g(2x + 5) 3 B ...

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• [PDF File]Math 1A: Homework 7 Solutions

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  (a;b). Note however that f0(x) = 3x 2+ ex>0 for all xsince x 0 and ex>0. This shows that f0(d) cannot be zero. We therefore have a contradiction, showing that f(x) = 0 has at most one solution in the (1 ;1). Couple this with the fact that f(x) = 0 has at least one solution to conclude that this equation has exactly one solution in the given ...

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• [PDF File]Math 112 (Calculus I) Final Exam Form A KEY

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  a) f has an inflection point at x = 2. b) f is increasing in a neighborhood around x = 2. c) f has a local minimum at x = 2. d) f has a local maximum at x = 2. e) f is decreasing in a neighborhood around x = 2. f) we don’t have enough information to prove that any of these are true. Solution: c)

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• [PDF File]Homework 5 Solutions.

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  Homework 5 Solutions. x4.2 #1 d. Use the division algorithm to find the quotient and remainder when f(x) = 2x4 +x3 6x2 x+2 is divided by g(x) = 2x2 5 over Q. Solution: Long division gives:

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• [PDF File]QUANTUM MECHANICS Examples of operators

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  A = d/dx, B = 3, f = f = x2 (d/dx +3) x2 = dx2/dx +3x2 = 2x + 3 x2 ABf = A (Bf) d/dx (3 x2) = 6x Note that A(Bf) is not necessarily equal to B(Af): A = d/dx, B = x, f = x2 A (Bf) = d/dx(x⋅ x2) = d/dx (x3) = 3 x2. B (Af) = x (d/dx x2) = 2 x2 In general, d/dx (xf) = f + x df/dx = (1 + x d/dx)f

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• [PDF File]Chapter 3: Practice/review problems

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  (a) 1/2 (b) 1 (c) 3/2 (d) 2 (e) 5/2 [29]. Consider the function f(x) = (2x2 +3 if x ≤ 3 3x+B if x > 3. Find a value of B such that f(x) is continuous at x = 3. (a) 6 (b) 9 (c) 12 (d) 15 (e) There is no such value of B. [30]. Find all values of a such that the function f(x) = ˆ x2 +2x if x < a −1 if x ≥ a is continuous everywhere.

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• [PDF File]SOLUTIONS - UCSD Mathematics | Home

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  Problem 2. Determine the global max and min of the function f(x;y) = x2 2x+2y2 2y+2xy over the compact region 1 x 1; 0 y 2: Solution: We look for the critical points in the interior:

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