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Find the intervals on which the function is increasing, decreasing, concave up, concave down, then find any local extreme values, and inflection point and explain.
y=-x^5+(7/3)x^3+5x^2+4x+2
This is a calculus problem.

2007-11-01 14:53:46 · 1 answers · asked by Anonymous in Science & Mathematics Mathematics

1 answers

Well, it's all about where the derivative is positive, negative, or zero. (And to a limited extent about the sign of the second derivative too.)

So let's look at the derivative. It's y' = 5x^4+ 7x^2 + 10x + 4.

Hmm. I suspect that function has no zeros at all. Certainly y' is positive when x >0 or x < -2 or so. It's also positive at x = -1.

Let's look at the second derivative. That would be 20x^3 + 14x + 10. That's negative for all x <-1 and positive for all x >0. So y' is decreasing until some point in the interval [-1,0]. So it's als non-zero except possibly somewhere in this interval.

Anyhow, I'm not going to pursue this to the end, but that's the plan of attack.

2007-11-03 02:17:35 · answer #1 · answered by Curt Monash 7 · 0 0

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