calc problem
- Thread starter mr.burger
- Start date
- Joined
- Apr 12, 2005
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This one sort of irritated me. This is a fundamental problem type. If you REALLY can't do this one, you need to go have a chat with your teacher.
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.
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.
Hello, mr.burger!
So you <u>are</u> in self-study . . . that explains your spectrum of questions.
. . . 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 critical values: . x = -2, -2/3
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
So you <u>are</u> in self-study . . . that explains your spectrum of questions.
To find y-intercepts, let x = 0: . y .= .0<sup>3</sup> + 4·0<sup>2</sup> + 4·0 .= .0
. . . 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 critical values: . x = -2, -2/3
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
- 11,339
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:
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.