- Thread starter mr.burger
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We absolutely are not here to do your homework for you. It will be of no benefit to you if we do your section review for you.

So, what's it going to be? Where are you struggling, if anywhere? Pick a favorite problem or two and show us your work.

So you <u>are</u> in self-study . . . that explains your

To find y-intercepts, let x = 0: . y .= .0<sup>3</sup> + 4·0<sup>2</sup> + 4·0 .= .0Find the intercepts, local extrema and inflection points of y = x<sup>3</sup> + 4x<sup>2</sup> + 4x

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. . . The y-intercept is at: .(0,0)

To find x-intercepts, let y = 0: . x<sup>3</sup> + 4x<sup>2</sup> + 4x . = . 0

. . . We have: . x(x<sup>2</sup> + 4x + 4) .= .x(x + 2)<sup>2</sup> . = . 0

. . . Hence, we get: . x = 0, -2

. . . The x-intercepts are: . (0,0), (-2,0)

For extrema, set y' = 0 and solve: . y' .= .3x<sup>2</sup> + 8x + 4 . = . 0

. . . We get two

To test them, use y'' .= .6x + 8

When x = -2: . y'' .= .6(-2) + 8 .= .-4 . . . negative, concave down: ∩ . . . local maximum at (-2,0)

When x = -2/3: . y'' .= .6(-2/3) + 8 .= .+4 . . . positive, concave up: U . . . local minimum at (-2/3,-32/27)

For inflection points: solve y'' = 0

. . . We have: . 6x + 8 .= .0

. . . Hence, there is an inflection point at: .x = -4/3

- Joined
- Apr 12, 2005

- Messages
- 9,931

Maybe it isn't working? Did you start with the first section and ponder it until you got it? Generally, books are written in an appropriate sequential fashion.mr.burger said:I am just teaching myself calc and the book isn't helping me

It is POSSIBLE that one could skip all over and manage to learn something, but I think it would be a rare individual, indeed, who could succeed on that path.