# Finding the Inflection points of x^2-4√x

1. Apr 8, 2016

### AllanW

1. The problem statement, all variables and given/known data
find the inflection points of x^2-4√x

2. Relevant equations

3. The attempt at a solution
Okay, I started with finding the derivatives;
f'(x)=2x-2/√x
f''(x)=2+1/√x^3
and made the second derivative =0
(2+1/√x^3=0)(√x^3)
2√x^3+1=0
(√x^3=-1/2)^2
x^3=1/4
x=cube root(1/4)
x=0.63
But when I enter '0.63' into the second derivative i get '4' (2+1/(√0.63^3)=4) Should i not get 0? Does this mean there are no inflection points?

2. Apr 8, 2016

### andrewkirk

It's unclear what you are doing here:
Note that for the given expression x^2-4√x to be a function, we need to choose a branch of the square root. The convention is to choose the positive branch unless explicitly stated otherwise. Then when we get to (√x)^3=-1/2 we see that there is no solution, because in the positive branch (√x)^3>0. So there are no inflection points.

The solution you found is for the negative branch of the square root, which does have an inflection point.

3. Apr 9, 2016

### ehild

It was a real function, the domain is x>0. That means, the second derivative can not be zero. When you take the square of the equation, you introduce false root.

4. Apr 9, 2016

### Ray Vickson

You have $f(x) = f_1(x) + f_2(x)$, where $f_1(x) = x^2$ and $f_2(x) = -2 \sqrt{x}$. We have $f_1''(x)>0$ and $f_2''(x)>0$ on $\mathbb{R}_+ = \{ x > 0 \}$, so there are no inflection points. Basically, $f$ is the sum of two strictly convex functions on $\mathbb{R}_+$, so is itself strictly convex on that set.

5. Apr 10, 2016

### AllanW

great, thanks