A Model Solution and More

1. A Model Solution and More

2. Sketch the graph of y = Y- intercepts: For y-intercepts, set x = 0 The y-intercept is (0, -1) X- intercepts: For X-intercepts, set y = 0 The x-intercept is (-1,0) Asymptotes x – 1 = 0 gives a restriction of x = 1 is a vertical asymptote. and

3. Asymptotes y = 1 is a horizontal asymptote. For critical points: For max/min points set y’ = 0 There are no critical points. But -2 ≠ 0

4. 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

5. Given: f’(x) = x4-4x3+2x2+4x-3 f”(x) = 4x3 – 12x2 + 4x + 4 For Critical Points: Set f’(x) = 0, using the factor theorem f’(x) = (x+1)(x3-5x2+7x-3) = (x+1)(x-1)(x2 - 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 -1 1 3 _ _ 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 an Inflection pt. at (1,6) since y’ < 0 for all x in (-1,1) and y’ < 0 for all x in (1,3). There is a local min. at (3,1.5) since y’ < 0 for all x in (1,3) and y’ > 0 for all x in (3,∞) .

6. 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.

7. Sketch the Graph of y = f(x) given the following information:

8. Sketch the Graph of y = f(x) given the following information:

9. Sketch the Graph of y = f(x) given the following information:

10. Sketch the Graph of y = f(x) given the following information: