Why is x = -1 not an inflection point of y = (1/3) (x^2 - 1)^{2/3} ?

Barry75

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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
 

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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

First, you'll want to state the definition you have been given, because I think different sources give slightly different definitions. For example, Wikipedia says that the function must be differentiable at the point. What does your textbook say?

But, second, when I graph it, the second derivative appears to be negative on both sides of x=-1. Please check your work on that. If necessary, show us your work so we can check it. But, for example, when x = -1.01, I get (using your formula) y" = -102.93.
 
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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

what does inflection point mean?
and; what values do the first and/or second derivative have at the inflection point?
 
Last edited by a moderator:
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
Please factor f ''(x) completely and we will see if it does have a sign change at x = -1 as you change. If what you wrote above is correct, then yes there is a POI at x = -1
 
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Thanks all... I think I figured out my issue. I was about to post all my work but it clicked. I was using f''(-2) and f''(0) to check whether the values on both sides of f''(-1) changed sign. I should have checked values closer to -1 as I didn't have the full picture of all critical values and candidates of inflection points. When I plugged in -1.1 as Dr.Peterson indicated, I got negatives on both sides. But thank you for confirming I was along the right path.

Thanks Again,
Barry
 
Thanks all... I think I figured out my issue. I was about to post all my work but it clicked. I was using f''(-2) and f''(0) to check whether the values on both sides of f''(-1) changed sign. I should have checked values closer to -1 as I didn't have the full picture of all critical values and candidates of inflection points. When I plugged in -1.1 as Dr.Peterson indicated, I got negatives on both sides. But thank you for confirming I was along the right path.

Thanks Again,
Barry

As was suggested, factoring is commonly a better way to be sure of a sign change, rather than hoping you chose numbers close enough. You have f"(x) = (4/9)(x^2 - 1)^{-1/3} - (8/27)(x^2 - 1)^{-4/3}, so the lowest power of x^2 - 1 is -4/3; make this the common factor:

f"(x) = (4/9)(x^2 - 1)(x^2 - 1)^{-4/3} - (8/27)(x^2 - 1)^{-4/3}
= [(4/9)(x^2 - 1) - (8/27)]/(x^2 - 1)^{4/3}
= ((4/27)x^2 - (4/9))/(x^2 - 1)^{4/3}
= (4/27)(x^2 - 3)/(x^2 - 1)^{4/3}.

This kind of factoring is very often useful in calculus.

The denominator is always positive; the numerator changes sign only at ±sqrt(3). So at ±1, where it is undefined, it doesn't change sign.
 
Thanks all... I think I figured out my issue. I was about to post all my work but it clicked. I was using f''(-2) and f''(0) to check whether the values on both sides of f''(-1) changed sign. I should have checked values closer to -1 as I didn't have the full picture of all critical values and candidates of inflection points. When I plugged in -1.1 as Dr.Peterson indicated, I got negatives on both sides. But thank you for confirming I was along the right path.

Thanks Again,
Barry
The same problem you had with what you first tried (using x=-2) can still be present with trying to use x=-1.1. You have no guaranty at all that if you try using x=-1.005 that the sign of f ''(-1.005) and f ''(-1.1) could be different. The only way to tell for sure is to completely factor f ''(x) and determine the sign of f'' if x is less than -1 and then the same when x is a little more than -1.
 
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