# Inflection points using 2nd derivative

1. Jul 8, 2008

### Beeorz

1. The problem statement, all variables and given/known data
Find all inflection points,

f(x) = -x^3/2

2. Relevant equations
n/a

3. 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??

2. Jul 8, 2008

### rock.freak667

I am assuming you meant that $f(x) = \frac{-x^3}{2}$

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. Jul 8, 2008

### Nick89

Judging by his derivatives, he actually means
$$f(x) = -x^{3/2}$$
which should have been written as -x^(3/2).

First of all, I don't think your derivatives are correct, check them again.

4. Jul 8, 2008

### Beeorz

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. Jul 8, 2008

### Feldoh

$$\frac{d}{dx}f(x) = \frac{-3}{2}x^{1/2}$$ 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. Jul 8, 2008

### Beeorz

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

7. Jul 8, 2008

### matt grime

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.