F x x2 3x x2 2x

    • [PDF File]How to find the equation of tangent line at a given point ...

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      (x x2) = 1 2x J Find the rst derivative of the function. m = f 0 (0) = 1 2(0) = 1 J Find the slope of the tangent line at the given point P. y (0) = 1[x (0)] J Use the Point-Slope formula: y y

    • [PDF File]Ms. Parnell's Class - Home

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      Perform the indicated operation. x2 x2 +3x—10 x2 +2x x2+3X+2 x2—6X+5 x2+X—6 x2+9X+14 x+7 (k+ò(k-2) (xyqì(xo) 10. 7y5 18X 21 3 x2 —13X+40

    • [PDF File]SOLUCIONES TEMA 6: DERIVADAS - LA CASA DE GAUSS

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      2.x o o 2x3 + 2x- 2x3 (x2 + Representa las siguientes funciones racionales, siguiendo los pasos de la pági- na anterior: x2 + 3x x2+2 x2+1 x x + 3x+ 11 (2x + 3) (x + l) _ (x2 + + 11) 2x2 + 2x + 3x + 3 — x x2 + 2.x — 8 11 —3, -1 0) Máximo en (—4, _5) Mínimo en (2, 7). Asíntota vertical: Asíntota oblicua: (2x + 3) (x + 1) — (x2 + 3x)

    • [PDF File]Graph Transformations - University of Utah

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      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.) 4 3x7 + 2 59 Exercises For #1-10, suppose f(x ...

    • [PDF File]Composite Functions - Practice (and solutions)

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      1 a) f(g(x)) 2(3x) + 3 = 6x + 3 b) g(f(x)) 3(2x + 3) 6x+ 9 d) g(g(x)) 3(3x) 9x b), 2 . b) d) g(g(x)) (x2 (x2) X 4 2X2 I X 4 I x3+X LEARNING COMMONS . Author: Jeffrey Heath Created Date:

    • Practice Fall Chapter - Houston Community College

      x2 - 2x - 15 x + 3 A) 5 B) 0 C) Does not exist D) -8 6) 7) lim ... x→∞ x2 + 3x + 4 x3 - 6x2 + 15 A) 0 B) 1 C) ∞ D) 4 15 14) 3. 15) lim x→-∞ ...

    • THE DIFFERENCE QUOTIENT - Whatcom Community College

      EXAMPLE 2: A Quadratic Expression f(x)=x2+2x fxhfx Find h +− Let’s begin with the two rooms in the top. f(x+h)−f(x) Withf(x)=x2+2xf(x+h)becomes(x+h)2+2(x+h) Simplifying, 2

    • [PDF File]Chapter R - Florida International University

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      vertex, axis of symmetry, y-intercept, and x-intercepts, if any. a) f (x) = 2x 2-x -1 b) f(x)=-x2 +2x-4 44. Determine, without graphing, whether the given quadratic function has the maximum value or the minimum value and then find this value. a) f(x)=2x2 -6x+l b) f(x)=-x2 -3x+5 45.

    • [PDF File]SOLUTIONS - University of California, San Diego

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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:

    • [PDF File]COLLEGE ALGEBRA REVIEW FOR TEST 3

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      55)a) f(x) = 7x + 4 2x - 1 b) f(x) = x - 6 x2 - 4 c) f(x) = 6x3 + 3x - 3 x2 + 4x - 21 Write a symbolic representation of a rational function f that satisfies the conditions. 56)Vertical asymptotes x = 3 and x = -8, horizontal asymptote y = 3 Sketch the graph of the rational function. 57)f(x) = x - 4 x + 5 58)f(x) = x2 - 16 x - 4 Solve the ...

    • [PDF File]Math 10007 Test 01 & Math 10023 Midterm Examination STUDY ...

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      x2 - 10x + 5 x2 + 2x - 1; x = 7 A)- 8 31 B)2 C)- 1 4 D) 8 31 1) Evaluate the rational function for the given replacement value. 2)f(x)= x2 - 10x + 5 x2 + 2x - 1; f( 7). A)Undefined B)- 8 31 C)2 D) 8 31 2) Evaluate the rational expression. 3)f(a)= a2 1 - a2; Find f(-9). A)- 80 81 B) 81 82 C)- 81 80 D) 81 80 3) Simplify the expression. 4) 3x + 2 ...

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

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      g(x)= x2 + 1 x (8) f(x) = 3x+ 4 (9) f( ) = 3( ) + 4 (10) f(g(x)) = 3(g(x)) + 4 (11) f(x2 + 1 x) = 3(x2 + 1 x) + 4 (12) f(x 2+ 1 x) = 3x + 3 x + 4 (13) Thus, (f g)(x) = f(g(x)) = 3x2 + 3 x + 4. Let’s try one more composition but this time with 3 functions. It’ll be exactly the same but with one extra step. Find (f g h)(x) given f, g, and h ...

    • [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 ...

    • [PDF File]EJERCICIOS RESUELTOS LÍMITES DE FUNCIONES - ManoloMat

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      X2 X2 2X+1 (X 1) Hana, si existen, estos limites. x2 2X+1 x 2 +4x—1 x2 6X+8 x3 -50 g) h) i) x x X4 im 3x 2x2 2x2 3x 2x2 j) k) x 1m x lim X4 3x 2x2 3x 2X2 3x 2x2 a) b) c) x-3 d) e) f) im 1m 1m 1m x x X x Ca cula IOS ímites en en de las funciones siguientes. b) f(x) a) E ímite en —T no existe pues a función soo está definida si x > 5 , im 2 2

    • [PDF File]Tarea 11 Ejercicios resueltos - CIMAT

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      3.Encuentre todos los valores de apara los que fes continua en R: f(x) = (x+ 1 si x a x2 si x>a Soluci on. f ser a continua s olo si l m x!a (x+ 1) = l m x!a+ x 2, es decir si a+ 1 = a2.Resolviendo esta ecuaci on obtenemos dos posibles soluciones:

    • [PDF File]EJERCICIOS POLINOMIOS Y FRACCIONES ALGEBRAICAS

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      x 2x—5 x2 + 3x —9 x2 + 6x + 9 029 Realiza las siguientes operaciones. 2 3x2 2 1 x 002 Multiplica estos polinomios. 1 1 001 Efectúa la siguiente operación. + + X 1)- (x3 + x2 1) 026 Simplifica estas fracciones algebraicas. —6 3 + 3X2 — 5x + 4 x 2 x 6x3 — 6x

    • [PDF File]Exam 1 Solutions

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      4. (10 points) Use the limit de nition to nd the derivative of f(x) = x2 2x. f0(x) = lim h!0 f(x+ h) f(x) h = lim h!0 (x+ h)2 2(x+ h) x2 2x h = lim h!0 x 2+ 2xh+ h 2x 2h x2 2x h = lim h!0 x2 + 2xh+ h2 2x 2h x2 + 2x h = lim h!0 2xh+ h2 2h h = lim h!0 h(2x+ h 2) h = lim h!0 (2x+ h 2) = 2x+ (0) 2 = 2x 2 5. (15 points) Use the ’shortcut rules ...

    • [PDF File]23 EJERCICIOS de DERIVADAS

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      x 1 y x + = b) 2x 3x+12 y x − = c) x 1 y 1 x + = − d) x2 y x = e) 3x 2x 54 2 y 2 − + = f) y 3x 5= +(2)5 g) 2 2x y x x 1 = + + (Sol: a) − 3 2 y'= x; b) − = 2 2 2x 1 y' x ; c) ( ) 2 = − 2 y' 1 x; d) = 3 x y' 2; e) − y'=6 x3 2; f) ( ) 4 y'=30x3x2 +5 g) ( ) − = 2 2 2 2x +2 y' x+1) 13. Hallar la fórmula para la derivada de e ...

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