Inflection points using 2nd derivative

So there are no inflection points.In summary, the function f(x) = -x^(3/2) does not have any inflection points. This is because the second derivative, f''(x) = (-3/4)x^(-1/2), is not defined at x=0 and there is no change in sign when testing points to the left and right of x=0.
  • #1
Beeorz
30
0

Homework Statement


Find all inflection points,

f(x) = -x^3/2


Homework Equations


n/a



The Attempt at a Solution


f'(x) = -3/2x^-1/2
f''(x) = 3/4x^-3/2

The f''(x) does not exist. I know that doesn't necessarily mean that there isn't any inflection points, but how do I go about finding them? Ususally I just set the 2nd derivate equal to 0 but it doesn't seem to work in this case. When I do set it equal to 0, I get x=0. And when I check points to the left and right of f''(0) the 2nd derivate doesn't change signs. Does that mean there is no inflection points??
 
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  • #2
I am assuming you meant that [itex]f(x) = \frac{-x^3}{2}[/itex]


But you are correct in saying that f''(x)=0 => x=0 . f(0)=?

Now check the sign of f'(x) at say x=0.1 and at x=-0.1. See if the sign changes.

[You are to check if the sign of the gradient changes which is given by f'(x)]
 
  • #3
Judging by his derivatives, he actually means
[tex]f(x) = -x^{3/2}[/tex]
which should have been written as -x^(3/2).

First of all, I don't think your derivatives are correct, check them again.
 
  • #4
Yes, I meant f(x) = -x^(3/2)

wish I knew how to use w/e program that you guys are using so that there is less confusion..in looking at my derivative..they seem correct to me

basically -(3/2)x^((3/2)-1) = -(3/2)x^(1/2)...no?

ill check over them again and see
 
  • #5
[tex]\frac{d}{dx}f(x) = \frac{-3}{2}x^{1/2}[/tex] is correct, however that is not what you had in your first post.

However there cannot be a point of inflection at x=0 because the domain is only defined from [0,inf) which automatically means that the second derivative cannot change signs at x=0 because it to is only defined on (0,inf)
 
  • #6
Ah, I see my mistake.So,

f'(x) = (-3/2)x^(1/2)
f''(x) = (-3/4)x^(-1/2)

(-3/4)x^(-1/2)=0
x=0

Taking test points to the left and right of x=0 yields no sign change, thus there is no inflection point.

Would what I said above be correct to say? We currently solve for inflection points setting the 2nd derivative equal to zero. We then use those zeros and divide the x-axis into intervals. Using a test point within those intervals we determine the sign of f''. If f'' changes sign then there is a inflection point.

Just wanting a confirmation :smile:
 
  • #7
You can't 'test' to the left of zero, i.e. negative numbers. As has been pointed out, this function isn't defined for negative x, and you just took a square root of a negative number in your last post as well in the higher derivatives.
 

What is an inflection point?

An inflection point is a point on a graph where the curvature changes, indicating a shift in the direction of the curve.

How is an inflection point identified using the 2nd derivative?

An inflection point can be identified by finding the second derivative of a function and setting it equal to zero. The x-value that satisfies this equation is the inflection point.

What information can be gathered from inflection points?

Inflection points can provide information about the concavity of a curve, as well as the direction of the curve at that point.

How are inflection points used in real-world applications?

Inflection points are used in a variety of fields, such as economics, engineering, and physics, to analyze trends and make predictions about future behavior.

Can a function have multiple inflection points?

Yes, a function can have multiple inflection points. These points can occur when there are changes in the concavity of the curve at different points.

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