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critical points |
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| Mar13-05, 03:17 PM | #1 |
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critical points
hi guys i'm stuck on this, i keep thinking i have the right answeres but i can't get them.
the function is f(x): -2x^3 + 39x^2 - 180x + 1 i need to list all of the critical points, and hten indicate where it is increasing and where it is decreasing. I set the derivative -6x^2 + 78x - 180 = 0 and then solved for x. this is correct right. by doing so i ended up with critical points at -2 and 15. Then made a number line and figured that the slope before -2 was neg, b/w -2 and 15 it was pos, and after 15 it was neg. so my interval of increasing was (-2,15) and the interval of decreasing was (-inf, -2) U (15,inf) where am i going wrong - PLEASE HELP ME |
| Mar13-05, 03:41 PM | #2 |
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Did u compute the second derivative,too...?To find the nature of those 2 points,u must evaluate the second derivative in those points.
Daniel. |
| Mar13-05, 03:43 PM | #3 |
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Recognitions:
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You made a mistake in solving the quadratic. The roots are 3 and 15.
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| Mar13-05, 03:44 PM | #4 |
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critical points
Yes i determined the second derivative as -12x + 78. Then using the second derivative rule plugged in 15, and -2 telling me that -2 was a local min and 15 was a local max. DO you know if 15, -2 are the only critical points in my problem????
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| Mar13-05, 03:49 PM | #5 |
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i don't see how u got 3 and 15 ??
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| Mar13-05, 03:58 PM | #6 |
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The roots are 3 & 10.
Daniel. |
| Mar13-05, 03:59 PM | #7 |
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-6x^2 + 78x - 180 = 0
x^2 -13x + 30 = 0 x=(13+/-sqrt(169-4*1*30))/2 x=(13+/-7)/2 x=20/2 = 10 x=6/2= 3 |
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