y' = 3x^2 + 12x + 9 = 3(x^2 + 4x + 3) = 3(x+1)(x+3), so critical numbers are x = -1, -3
y'' = 3(2x+4) = 6(x+2), so inflection point is at x = -2.
at x = -1, y'' > 0, so x = -1 is a min
at x = -3, y'' < 0, so x = -3 is a max
y is increasing for x < -3 and x > -1, and decreasing otherwise
y is concave up for x > -2, and concave down for x < -2
2006-11-02 16:04:52
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answer #1
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answered by James L 5
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y' = 3x^2 + 12x + 9 = 3(x^2 + 4x + 3) = 3(x+1)(x+3),
so critical points are x = -1, -3
y'' = 3(2x+4) = 6(x+2), so inflection the point is x = -2.
at x = -1, y'' > 0, so x = -1 is a min
at x = -3, y'' < 0, so x = -3 is a max
y is increasing for x < -3 and x > -1, and decreasing otherwise
y is concave up for x > -2, and concave down for x < -2
s
2006-11-03 14:52:26
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answer #2
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answered by Anonymous
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Compute its derivative to find the answers.
y=x^3+6x^2+9x,
so y'=3x^2+12x^2+9, and y''=6x+12.
let y'=0, we can find x=-1 or x=-3.
when x=-1,y''>0,white y''<0 when x=-3.
So, you can make analysis based on these two points.
2006-11-02 16:24:13
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answer #3
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answered by Anonymous
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don't know how to sketch it on the computer..
(-3,0) touches graph and comes back down parabola opens down
(-1,4)mini parabola opens up
(0,0) goes through origin
skims up along the y-axis
Range =-infinity to infinity
kind of makes a shape like an "N" but not quite.
2006-11-02 16:07:06
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answer #4
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answered by Dawn J 4
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Yeah it would make experience in case you needed help with only one difficulty to comprehend the basics, yet you % people to do all your homework. attempt no longer dozing by type next time.
2016-11-27 00:41:53
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answer #5
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answered by Anonymous
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