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

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Homework Statement


find the inflection points of x^2-4√x

Homework Equations

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?
 
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It's unclear what you are doing here:
AllanW said:
2√x^3+1=0
(√x^3=-1/2)^2
x^3=1/4

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.
 
AllanW said:

Homework Statement


find the inflection points of x^2-4√x

Homework Equations

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)

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.
AllanW said:
(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?
 
AllanW said:

Homework Statement


find the inflection points of x^2-4√x

Homework Equations

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?

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.
 
Ray Vickson said:
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.
great, thanks
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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