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-3 to -1 seems to be downwards(concave) so I chose that
1 to 2 is also downward(concave)
and 2 to ∞ is downwards
Am I right?
Last edited by Mathegocart (2016-01-27 16:00:06)
The integral of hope is reality.
May bobbym have a wonderful time in the pearly gates of heaven.
He will be sorely missed.
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hi mathgogocart
I had to look up concavity as I'd not met it before. Here's a link:
https://www.math.hmc.edu/calculus/tutor … condderiv/
You graph may be divided into 5 sections: (1) up to -3 (2) -3 to -1 (3) -1 to +1 (4) +1 to +2 (5) above + 2
From that page it would seem that the function is concave upwards for (2) and (4) and concave downwards elsewhere. Hope that helps,
Bob
ps. The theorem only applies to open intervals as f'' is zero at the 4 points.
pps. If you attempt a sketch of the f' graph you will see that, for example, in the interval -3 to -1, the gradient is increasing meaning the f graph is concave upwards.
Children are not defined by school ...........The Fonz
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hi mathgogocart
I had to look up concavity as I'd not met it before. Here's a link:
https://www.math.hmc.edu/calculus/tutor … condderiv/
You graph may be divided into 5 sections: (1) up to -3 (2) -3 to -1 (3) -1 to +1 (4) +1 to +2 (5) above + 2
From that page it would seem that the function is concave upwards for (2) and (4) and concave downwards elsewhere. Hope that helps,
Bob
ps. The theorem only applies to open intervals as f'' is zero at the 4 points.
pps. If you attempt a sketch of the f' graph you will see that, for example, in the interval -3 to -1, the gradient is increasing meaning the f graph is concave upwards.
Thank you!
The integral of hope is reality.
May bobbym have a wonderful time in the pearly gates of heaven.
He will be sorely missed.
Offline
Pages: 1