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A Model Solution and MorePowerPoint Presentation

A Model Solution and More

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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

y = 1 is a horizontal asymptote.

For critical points:

For max/min points set y’ = 0

There are no critical points.

But -2 ≠ 0

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

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,∞) .

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.

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

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

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

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

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