Christos
f(x) = (2x +1)^3
f'(x) = 6(2x + 1)^2
f''(x) = 48x + 24
I need to know when its concave up/down increasing /decreasing and the inflection points I am new to this kind of stuff
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rulnick
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First derivative is nonnegative for all real x, so f is non-decreasing.
Second derivative is everywhere matching the sign of x+1/2, so there is an inflection point at x=-1/2. The function is concave down on x<-1/2 and concave up on x>-1/2.
Christos
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how did you find the -1/2
rulnick
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f''(x)=0 at x=-1/2
Christos
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Ok and something more
are my derivative calculations correct?
f(x) = (2x +1)^3
f'(x) = 6(2x + 1)^2
rulnick
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Yes, all were perfect!
Christos
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so its not decreasing that means its always increasing? Kinda what's the interval?
Christos
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(0,infinity) increasing?
rulnick
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non-decreasing means increasing or flat. it is flat at the inflection point, increasing everywhere else
rulnick
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so increasing on the entire real line except at -1/2, where it is flat (deriv=0)
Christos
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here it asks me the open interval on which f is increasing what should I put? (-inf,-1/2)U(-1/2,int) ?
rulnick
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yes, very nicely done
Christos
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and decreasing interval*
rulnick
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empty set
Christos
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like I just say "it's not decreasing anywhere" ?
rulnick
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yes
Christos
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Alright, thank you!
rulnick
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welcome