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Published byElwin Charles Modified over 9 years ago
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A Model Solution and More
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Sketch the graph of y = Y- intercepts: For y-intercepts, set x = 0 X- intercepts: For X-intercepts, set y = 0 The x-intercept is (-1,0) The y-intercept is (0, -1) Asymptotes and x – 1 = 0 gives a restriction of x = 1 is a vertical asymptote.
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Asymptotes y = 1 is a horizontal asymptote. For critical points: For max/min points set y’ = 0 But -2 ≠ 0 There are no critical points.
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Increasing/Decreasing Regions For increasing regions, y’>0 For decreasing regions, y’<0 < 0, for all x, x ≠1, the curve is always decreasing For Inflection Points: Check y” = 0 y” ≠ 0 for all x, x ≠ 1 there are no inflection points
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For Critical Points: Set f’(x) = 0, using the factor theorem f’(x) = (x+1)(x 3 -5x 2 +7x-3) = (x+1)(x-1)(x 2 - 4x+3) = (x+1)(x-1) 2 (x-3) there are critical points at x = 1, -1, 3 For Max/Min: examine sign of f’(x) near the critical points Given: f’(x) = x 4 -4x 3 +2x 2 +4x-3 f”(x) = 4x 3 – 12x 2 + 4x + 4 13 Sign of f’(x)+ _ + _ There is a local max. at (-1,10) since y’ > 0 for all x in (-∞,-1) and y’ < 0 for all x in (-1,1). There is a local min. at (3,1.5) since y’ 0 for all x in (3,∞). There is an Inflection pt. at (1,6) since y’ < 0 for all x in (-1,1) and y’ < 0 for all x in (1,3).
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Concavity using f ’’(x) f ”(-1) = -16, since f ”(x) < 0, therefore a local max f ”(1) = 0, since f ”(x) = 0, therefore not concave, suspect an inflection point –> check signs: since f ” > 0 for all x in (-1,1) and f ” < 0 for all x in (1,3) f ”(3) = 16, since f ”(x) > 0, therefore a local min There are no vertical asymptotes For Horizontal asymptotes – since is the dominant term in f(x), the function will tend towards y = as the end behaviour.
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xx<111<x<222<x<33x>3 f(x)-4-2-2 0 f’(x)-0+++0- f”(x)+++0--- Sketch the Graph of y = f(x) given the following information:
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xx<0.50.50.5<x<111<x<22x>2 f(x)0 f’(x)-0+++0+ f”(x)+++0-0+ Sketch the Graph of y = f(x) given the following information:
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x--0 f(x)-0 f’(x)-0+++0+++0- f”(x)+++0-0+0---
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Sketch the Graph of y = f(x) given the following information: x11.523.24 f(x)57.710-7.9-22 f’(x)-0+++0---0+ f”(x)+++0---0+++
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