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concavity from the first derivative?
- Thread starter Cuddles
- Start date
Yep. Just use your second derivatives test. Note that even though you don't know exactly what h(x) is due to the unknown constant that arises from integration of h'(x), it wouldn't matter as this constant only affects the vertical shifts of h(x). The characteristics it would have (critical points and whatnot) would still be located at their corresponding x-values no matter what the constant of integration is. So yes, you can still determine the nature of h(x)'s concavity given h'(x).
Hello, Cuddles!
Yes, there is . . .
You found that \(\displaystyle x = \sqrt{2}\) is a critical value, right?
The slope there is zero (horizontal tangent).
Test a value to the left of the critical value.
. . \(\displaystyle \text{Say, }x = 1:\;\;h'(1) \:=\:\frac{1^2-2}{1} \:=\:\text{negative slope}\;\searrow\)
Test a value to the right of the critical value.
. . \(\displaystyle \text{Say, }x = 2:\;\;h'(2) \:=\:\frac{2^2-2}{2} \:=\:\text{positive slope}\;\nearrow\)
Now we know the shape of the curve near that critical value: .\(\displaystyle \searrow\;_{\rightarrow}\;\nearrow\)
Therefore, the critical value is a minimum.
\(\displaystyle \text{Is there a way to find the concavity from the first derivative?}\)
\(\displaystyle \text{For example: }\;h'(x) \:=\:\frac{x^2 -2}{x}\)
Yes, there is . . .
You found that \(\displaystyle x = \sqrt{2}\) is a critical value, right?
The slope there is zero (horizontal tangent).
Test a value to the left of the critical value.
. . \(\displaystyle \text{Say, }x = 1:\;\;h'(1) \:=\:\frac{1^2-2}{1} \:=\:\text{negative slope}\;\searrow\)
Test a value to the right of the critical value.
. . \(\displaystyle \text{Say, }x = 2:\;\;h'(2) \:=\:\frac{2^2-2}{2} \:=\:\text{positive slope}\;\nearrow\)
Now we know the shape of the curve near that critical value: .\(\displaystyle \searrow\;_{\rightarrow}\;\nearrow\)
Therefore, the critical value is a minimum.