A Model Solution and More

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.

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.

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

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 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).

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.

xx<111<x<222<x<33x>3 f(x) f’(x) f”(x) Sketch the Graph of y = f(x) given the following information:

xx< <x<111<x<22x>2 f(x)0 f’(x) f”(x) Sketch the Graph of y = f(x) given the following information:

x--0 f(x)-0 f’(x) f”(x)

Sketch the Graph of y = f(x) given the following information: x f(x) f’(x) f”(x)