Hi,
I have an equation (see below) and I'm being asked if there is an inflection point at x = -1.
. . . . .\(\displaystyle y\, =\, \dfrac{1}{3}\,\left(x^2\, -\, 1\right)^{\frac{2}{3}}\)
. . . . .\(\displaystyle y'\, =\, \left(\dfrac{4}{9}\right)\, x\, (x^2\, -\, 1)^{-\frac{1}{3}}\)
. . . . .\(\displaystyle y''\, =\, \left(\dfrac{4}{9}\right)\, (x^2\, -\, 1)^{-\frac{1}{3}}\, -\, \left(\dfrac{8}{27}\right)\, (x^2)\, (x^2\, -\, 1)^{-\frac{4}{3}}\)
Based on the answer key, it is not but it doesn't explain why except it does indicate first derivative is undefined but I didn't think that mattered as long as f(x) is continuous. As I checked the following conditions can someone tell me why this is not an inflection point?
1. f(x) is continuous at -1
2. Second derivative is undefined at -1
3. Second derivative of f(x) is positive when x < -1
4. Second derivative of f(x) is negative when x > -1
Thanks,
Barry
I have an equation (see below) and I'm being asked if there is an inflection point at x = -1.
. . . . .\(\displaystyle y\, =\, \dfrac{1}{3}\,\left(x^2\, -\, 1\right)^{\frac{2}{3}}\)
. . . . .\(\displaystyle y'\, =\, \left(\dfrac{4}{9}\right)\, x\, (x^2\, -\, 1)^{-\frac{1}{3}}\)
. . . . .\(\displaystyle y''\, =\, \left(\dfrac{4}{9}\right)\, (x^2\, -\, 1)^{-\frac{1}{3}}\, -\, \left(\dfrac{8}{27}\right)\, (x^2)\, (x^2\, -\, 1)^{-\frac{4}{3}}\)
Based on the answer key, it is not but it doesn't explain why except it does indicate first derivative is undefined but I didn't think that mattered as long as f(x) is continuous. As I checked the following conditions can someone tell me why this is not an inflection point?
1. f(x) is continuous at -1
2. Second derivative is undefined at -1
3. Second derivative of f(x) is positive when x < -1
4. Second derivative of f(x) is negative when x > -1
Thanks,
Barry
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