8 x sqrt 1 2x

• [PDF File]Finding Square Roots Using Newton’s Method

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  Example: To 20 decimal places, √ 7 = 2.6457513110645905905. Let’s see what Newton’s method gives with the initial approximation x0 = 3: x1 = 2.6666666666666666666 x2 = 2.6458333333333333333 x3 = 2.6457513123359580052 x4 = 2.6457513110645905908 Remarkable accuracy.

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• Finding the Equation of a Tangent Line

  -3 -2 -1 1 2 3 1) Find the first derivative of f(x). 2) Plug x value of the indicated point into f '(x) to find the slope at x. 3) Plug x value into f(x) to find the y coordinate of the tangent point.

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• [PDF File]Graph Transformations - University of Utah

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  1.) f(x) + 2 . A.) (x)8 2.) 3f(x) B.) 1 3 x 8 3.) f(x) C.) x8 ( 2 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 ...

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• [PDF File]Instantaneous Rate of Change — Lecture 8. The Derivative.

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  Example 1. Let f(x) = 1/x and let’s find the instantaneous rate of change of f at x 0 = 2. The first step is to compute the average rate of change over some interval x 0 = 2 to x; and in order for this to make sense we need x 6= 2. So that average rate of change is ∆y ∆x = f(2)−f(x) 2−x = 1 2 − 1 x 2−x = x−2 2x(2−x) = − 1 2x.

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• [PDF File]Approximating functions by Taylor Polynomials.

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  8 = 1 + x + x2 2! + x3 3! +···+ x8 8! The nth Taylor Polynomial for sinx for x near a = 0. First calculate the derivatives of sin x! You should find a pattern that makes this easy. derivative at x = 0 f (x) = sinx is 0 f (x) = cosx is 1 f (x) = f (3)(x) = f (4)(x) = f (5)(x) = f (6)(x) = Now put it together: sinx ≈ P n(x) = 0 + x + 0 2 ...

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• [PDF File]Working a difference quotient involving a square root

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  x+h x h = p x+h x h p p x+h+ x p x+: The key idea is that the numerators multiply in a nice way. Note that the two numerators together have the form (A B)(A+B) which is equal to A2 B2 (you might recall the phrase difference of squares). The squaring eliminates the square roots from the numerator. As a result, our expression above becomes p x+h ...

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• [PDF File]Di erentiation - Past Edexcel Exam Questions

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  Given that y = 5x4 3 + 2x2 x 3 p 2 x, x > 0, (b) nd dy dx, simplifying the coe cient of each term. [4] 32. (Question 11 - C1 January 2009) The curve C has equation y = 9 4x 8 x; x > 0: The point P on C has x-coordinate equal to 2. (a) Show that the equation of the tangent to C at the point P is y = 1 2x. [6]

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• [PDF File]Solution to Math 2433 Calculus III Term Exam. #3

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  1 3 x3 + 3x2 1 4 x4 = 9 + 27 81 4 + 8 3 12 + 4 = 125 12 2. Let T be the solid bounded by the paraboloid z= 4 x2 y2 and below by the xy-plane. Find the volume of T. (Hint, use polar coordinates). Answer The intersection of z= 4 2x 22y and xyplane is 0 = 4 x2 y;i.e. x2 +y = 4: In polar coordinates, ...

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• [PDF File]Math 2260 Exam #1 Practice Problem Solutions

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  Also, the points of intersection occur when 2x+ 7 = x2 1 or, equivalently, when 0 = x2 2x 8 = (x 4)(x+ 2); so the curves intersect when x= 4 and x= 2. Therefore, integrating top minus bottom over this region should yield the area between the curves: Z 4 2 (2x+ 7) x2 1 dx= Z 4 2 2x+ 8 x2 dx = x2 + 8x x3 3 4 2 = 16 + 32 64 3 4 16 + 16 3 = 48 64 3 ...

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• [PDF File]Techniques of Integration - Whitman College

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  1− x2, but it is not apparent that the chain rule is involved. Some clever rearrangement reveals that it is: Z x3 p 1− x2 dx = Z (−2x) − 1 2 (1−(1−x2)) p 1− x2 dx. This looks messy, but we do now have something that looks like the result of the chain rule: the function 1 − x2 has been substituted into −(1/2)(1 − x) √ x ...

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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]Math 2250 Exam #1 Practice Problems

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  8.Find the one-sided limit lim x! 1 x 1 x4 1 Answer: Notice that, as x! 1, the numerator goes to 2, while the denominator goes to zero. Hence, we would expect the limit to be in nite. However, it could be either 1 or +1, so we need to check the sign of the denominator. When x< 41, the quantity x >1, so x4 1 >0:

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• [PDF File]Volumes by Cylindrical Shells: the Shell Method

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  region bounded by y = 8 − x2, y = 2x, and the y-axis, about the y-axis. The upper curve is the parabola y = 8 − x2, and y = 2 x is the lower curve. Their first quadrant intersection is at (2, 4), therefore, V 2 x((8 x) 2x)dx 2 x(8 2x x)dx 2 (8 x 2x x3)dx 2 0 2 2 0 2 2 0

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• [PDF File]MATH 1B—SOLUTION SET FOR CHAPTERS 8.1, 8

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  MATH 1B—SOLUTION SET FOR CHAPTERS 8.1, 8.2 Problem 8.1.1. Use the arc length formula to find the length of the curve y = 2 − 3x,−2 ≤ x ≤ 1. Check your answer by noting that the curve is a line segment and calculating its length by the distance formula. Solution. First, note: y0 = −3 q 1+(y0)2 = √ 10

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• [PDF File]Integration by substitution

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  3. Finding Z f(g(x))g′(x)dx by substituting u = g(x) Example Suppose now we wish to find the integral Z 2x √ 1+x2 dx (3) In this example we make the substitution u = 1+x2, in order to simplify the square-root term. We shall see that the rest of the integrand, 2xdx, will be taken care of automatically in the

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• [PDF File]Solids with Known Cross-Sections Solutions - Weebly

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  axis, the y-axis, and the line x + 2 y = 8, as shown in the figure. If cross sections of the solid perpendicular to the x-axis are semicircles, what is the volume of the solid? —0.5 -1.0 c. 6.755 A. 12.566 D. 67.021 B. 14.661 E. 134.041 The region bounded by the graph of y = 2x —x2 and the x-axis is the base of a solid. For this solid,

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• [PDF File]Functions and abstraction

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  def dbl_plus(x): return 2 * x + 1 Defining a function Keyword that means: I am defining a function Keyword that means: This is the result Input variable name, or “formal parameter” Name of the function. Like “y= 5” for a variable 2x + 1 x Return expression (part of the returnstatement) 7

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• [PDF File]TRIGONOMETRY LAWS AND IDENTITIES

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  1+tan(x)tan(y) LAW OF SINES sin(A) a = sin(B) b = sin(C) c DOUBLE-ANGLE IDENTITIES sin(2x)=2sin(x)cos(x) cos(2x) = cos2(x)sin2(x) = 2cos2(x)1 =12sin2(x) tan(2x)= 2tan(x) 1 2tan (x) HALF-ANGLE IDENTITIES sin ⇣x 2 ⌘ = ± r 1cos(x) 2 cos ⇣x 2 ⌘ = ± r 1+cos(x) 2 tan ⇣x 2 ⌘ = ± s 1cos(x) 1+cos(x) PRODUCT TO SUM IDENTITIES sin(x)sin(y ...

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• [PDF File]Quiz 8 - University of California, Berkeley

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  x, so our normal methods do not apply. There are two ways to compute this derivative (both substantially the same). The rst is by logarithmic di erentiation. Write y = (x2 1)sinx and take the logarithm of both sides to obtain lny = sin(x)ln(x2 1). Then, we implicitly di erentiate each side, 1 y dy dx = sin(x) 2x x2 1 + cos(x)ln(x2 1); and ...

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• [PDF File]Techniques of Integration - Whitman College

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  204 Chapter 10 Techniques of Integration EXAMPLE 10.1.2 Evaluate Z sin6 xdx. Use sin2 x = (1 − cos(2x))/2 to rewrite the function: Z sin6 xdx = Z (sin2 x)3 dx = Z (1− cos2x)3 8 dx = 1 8 Z 1−3cos2x+3cos2 2x− cos3 2xdx. Now we have four integrals to evaluate: Z 1dx = x and Z

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