Curve Sketching. MATH 1310 Lecture 26 1 of 14 Ronald Brent 2016 All rights reserved.
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1 Curve Sketching 1. Domain. Intercepts. Symmetry. Asymptotes 5. Intervals of Increase or Decrease 6. Local Maimum and Minimum Values 7. Concavity and Points of Inflection 8. Sketch the curve MATH 110 Lecture 6 1 of 1 Ronald Brent 016 All rights reserved.
2 1 Using these steps let s analyze the function f 1. Domain: Since f is a polynomial,, is the domain. Intercepts: y-intercept is 0, 1 intercepts come from solving 1 0, Using w gives 1 w w 1 0 w, and so. Symmetry: Since f f, the function is symmetric w.r.t. y- ais. Asymptotes: None Since it s a polynomial. MATH 110 Lecture 6 of 1 Ronald Brent 016 All rights reserved.
3 5 Intervals of Increase and Decrease. Here f, and f. We can factor the first derivative as f and diagram its sign below Sign of f f is increasing on the intervals,0 and, and decreasing on, and 0,.,0, are critical points. 6 Relative Etrema: From the diagram f are relative minimum values and f 0 1 is a relative maimum value. MATH 110 Lecture 6 of 1 Ronald Brent 016 All rights reserved.
4 7 Concavity: Diagramming the second derivative f : Sign of f Concave up on the intervals, and, Concave down on the interval, Since the concavity changes at, there are inflection points there. Note; the actual point of inflection is 11, 9 MATH 110 Lecture 6 of 1 Ronald Brent 016 All rights reserved.
5 f Graph: y f MATH 110 Lecture 6 5 of 1 Ronald Brent 016 All rights reserved.
6 Using these steps let s eamine the function f. 1 Domain: This function is defined for 1. Intercepts: y-intercept is 0, -intercept is 0, 0. Symmetry: Since f f and f f the function has no typical symmetry. Asymptotes: lim f 0 so, y = 0 is a horizontal asymptote, lim f 1 asymptote. lim 1 f so = 1 is a vertical MATH 110 Lecture 6 6 of 1 Ronald Brent 016 All rights reserved.
7 5 Intervals of Increase and Decrease. f Critical Numbers are 0,1, Sign of f So f is decreasing on, 0, 1,, and increasing on, 0. 6 It s pretty clear that there is a relative minimum value at, the value / is And a relative maimum value of 0, at 0 MATH 110 Lecture 6 7 of 1 Ronald Brent 016 All rights reserved.
8 MATH 110 Lecture 6 8 of 1 Ronald Brent 016 All rights reserved. 7 Working with f. f To eamine concavity and points of inflection we need find solution to Using the substitution w, we re really solving a quadratic equation w w, which has solutions:
9 w , Now the values are w c, c c c Sign of f So the function f is concave up on c, c 1,, and concave down on 1, c,. c 1 Points of inflections are 7 5, MATH 110 Lecture 6 9 of 1 Ronald Brent 016 All rights reserved.
10 y MATH 110 Lecture 6 10 of 1 Ronald Brent 016 All rights reserved.
11 Eample: graph it. Let f e. Find out everything about this function and then Domain:, Intercepts: y-intercept is 0, 0 intercepts come from solving e 0, so 0. Symmetry: No Symmetry Asymptotes: Since lim f 0, there is a horizontal asymptote as approaches infinity. 5,6 Intervals of Increase and Decrease. Rel. Etrema Since f e e 1 e, 1 is a stationary point. Also, f 0 for 1, and f 0 for 1. So f increases on 1, and 1 decreases on 1. The is a relative ma at 1, e MATH 110 Lecture 6 11 of 1 Ronald Brent 016 All rights reserved.
12 7,8 Since f e 1 e e, is an inflection point, because f 0 for, and f 0 for. So f is concave up on, and concave down on. The function looks like: y MATH 110 Lecture 6 1 of 1 Ronald Brent 016 All rights reserved.
13 Let f ln. Find out everything about this function and then graph it. Domain: 0 Intercepts: Note: lim ln 0. 0 Symmetry: No Symmetry intercepts come from solving ln 0, so 1. Asymptotes: Since lim f, there are no asymptotes. 5,6 Intervals of Increase and Decrease. Rel. Etrema 1 Since f ln ln, e 1/ is a stationary point. Also, f 0 for e 1/ 1/, and f 0 for e. So f increases on 1/ e, and decreases on e 1/. The is a relative min at 1/ 1 1 e, e. MATH 110 Lecture 6 1 of 1 Ronald Brent 016 All rights reserved.
14 7,8 Since f ln ln, e / is an inflection point, because f 0 for / / e, and f 0 for e. So f is / / concave up on e, and concave down on e. The point of inflection is ^*ln e /, e..0 The function looks like: MATH 110 Lecture 6 1 of 1 Ronald Brent 016 All rights reserved.
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