calc problem

mr.burger

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Find the intercepts, local extrema and inflection points of y=x^(3)+4x^(2)+4x.
 

tkhunny

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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.
 

mr.burger

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Jun 12, 2005
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i am just teaching myself calc and the book isnt helping me so i just needed some help being walked through these problems. I am a visual learner.
 

soroban

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Hello, mr.burger!

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

Find the intercepts, local extrema and inflection points of y = x<sup>3</sup> + 4x<sup>2</sup> + 4x
.
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
 

tkhunny

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mr.burger said:
I am just teaching myself calc and the book isn't helping me
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.

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.
 
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