# Algebra step confusion and unnatural completing the square

• phospho
In summary, the solutions to the derivative of the function f(x) = ax-\frac{x^3}{1+x^2} have gone on to complete the square in an unnatural way.
phospho
Let $f(x) = ax - \dfrac{x^3}{1+x^2}$

where a is a constant
Show that, if a ≥ 9/8 then f'(x) ≥ 0

first problem

when taking the derivative in the solution they seem to have jumped a step which I don't see how:

$$f'(x) = a - \dfrac{3x^2(1+x^2) - 2x(x^3)}{(1+x^2)^2} = \dfrac{a + (2a - 3)x^2 + (a - 1)x^4}{(1 + x^2)^2}$$

Could anyone point out where the (2a - 3) came from, as well as the (a-1)

second problem,

the solutions have also go on to complete the square in a unnatural way (page 12 on link below), could anyone explain how they done this.

third problem, the solution also states the f(x) = -f(-x) so it's an odd function, how do we spot this exactly? Also, I thought that odd functions were in the form of f(-x) = -f(x)?

Thank you

Last edited by a moderator:
Your latex doesn't seem to work inside the box you provided can you edit it again.

Also you mention a link but there's none that I can see.

phospho said:
Let $f(x) = ax - \dfrac{x^3}{1+x^2}$

where a is a constant
Show that, if a ≥ 9/8 then f'(x) ≥ 0

first problem

when taking the derivative in the solution they seem to have jumped a step which I don't see how:

$$f'(x) = a - \dfrac{3x^2(1+x^2) - 2x(x^3)}{1+x^2)^2} = \dfrac{a + (2a - 3)x^2 + (a - 1)x^4}{(1 + x^2)^2}$$

Could anyone point out where the (2a - 3) came from, as well as the (a-1)

second problem,

the solutions have also go on to complete the square in a unnatural way (page 12 on link below), could anyone explain how they done this.

third problem, the solution also states the f(x) = -f(-x) so it's an odd function, how do we spot this exactly? Also, I thought that odd functions were in the form of f(-x) = -f(x)?

Thank you

Denoting ##\frac{d}{dx}## as D, we have
$$D \left(\frac{u}{v}\right) = \frac{Du}{v} - \frac{u \,Dv}{v^2}.$$
Apply this to u = x^3 and v = 1 + x^2. Then put everything over a common denominator.

Last edited:
Ray Vickson said:
Denoting ##\frac{d}{dx}## as D, we have
$$D \left(\frac{u}{v}\right) = \frac{Du}{v} - \frac{u \,Dv}{v^2}.$$
Apply this to u = x^3 and v = 1 + x^2. Then put everything over a common denominator.

I have no problem differentiating.

I have edited the post with the link

##f(x) = -f(-x)##

Multiply both sides by (-1)

##-f(x) = f(-x)## or ##f(-x) = -f(x)##

Karnage1993 said:
##f(x) = -f(-x)##

Multiply both sides by (-1)

##-f(x) = f(-x)## or ##f(-x) = -f(x)##

silly me, thanks

phospho said:
Let $f(x) = ax - \dfrac{x^3}{1+x^2}$

where a is a constant
Show that, if a ≥ 9/8 then f'(x) ≥ 0

first problem

when taking the derivative in the solution they seem to have jumped a step which I don't see how:

$$f'(x) = a - \dfrac{3x^2(1+x^2) - 2x(x^3)}{(1+x^2)^2} = \dfrac{a + (2a - 3)x^2 + (a - 1)x^4}{(1 + x^2)^2}$$

Could anyone point out where the (2a - 3) came from, as well as the (a-1)
First, get the denominator $(1+x^2)^2$ in the first term by mulitplying numerator and denominator by it.
$$\frac{a(1+ x^2)^2}{(1+ x^2)^2}- \frac{3x^2+ 3x^4- 2^4}{(1+ x^2)^2)}$$
Now multiply those out: $a(1+ x^2)^2= a(x^4+ 2x^2+ 1)= ax^4+ 2ax^2+ a$. Combine like terms: $$ax^4+ 2ax^2+ a- 3x^2+ 3x^4+ 2x^4= (a+3- 4)x^4+ (2a- 3)x^2+ a= (a- 1)x^4+ (2a- 3)x^2+ a$$ in the numerator.

second problem,

the solutions have also go on to complete the square in a unnatural way (page 12 on link below), could anyone explain how they done this.

third problem, the solution also states the f(x) = -f(-x) so it's an odd function, how do we spot this exactly? Also, I thought that odd functions were in the form of f(-x) = -f(x)?

Thank you

Last edited by a moderator:
HallsofIvy said:
First, get the denominator $(1+x^2)^2$ in the first term by mulitplying numerator and denominator by it.
$$\frac{a(1+ x^2)^2}{(1+ x^2)^2}- \frac{3x^2+ 3x^4- 2^4}{(1+ x^2)^2)}$$
Now multiply those out: $a(1+ x^2)^2= a(x^4+ 2x^2+ 1)= ax^4+ 2ax^2+ a$. Combine like terms: $$ax^4+ 2ax^2+ a- 3x^2+ 3x^4+ 2x^4= (a+3- 4)x^4+ (2a- 3)x^2+ a= (a- 1)x^4+ (2a- 3)x^2+ a$$ in the numerator.
argh so stupid, thank you!

I've managed to solve the problem by completing the square in the "natural" way, still not sure how they've then the unnatural way in the link in the original post.

## 1. What is completing the square?

Completing the square is a method used in algebra to solve quadratic equations or to convert a quadratic equation into its standard form. It involves manipulating the equation to create a perfect square trinomial, making it easier to solve or analyze.

## 2. Why is completing the square considered unnatural?

Completing the square can be considered unnatural because it involves adding and subtracting terms that are not originally present in the equation. This makes it a less intuitive method compared to other algebraic techniques.

## 3. Can completing the square be used to solve all quadratic equations?

Yes, completing the square can be used to solve any quadratic equation. However, it may not always be the most efficient method and other techniques such as factoring or using the quadratic formula may be more suitable.

## 4. How do I know when to use completing the square?

Completing the square is typically used when the quadratic equation does not have any obvious factors, making it difficult to solve using other methods. It is also commonly used when the equation is in standard form and needs to be converted into vertex form.

## 5. Can completing the square be applied to equations with higher degree terms?

No, completing the square is specifically used for quadratic equations, which have a degree of 2. It cannot be applied to equations with higher degree terms, such as cubic or quartic equations.

Replies
19
Views
2K
Replies
8
Views
973
Replies
10
Views
1K
Replies
2
Views
825
Replies
11
Views
1K
Replies
7
Views
1K
Replies
2
Views
1K
Replies
15
Views
1K
Replies
5
Views
1K
Replies
23
Views
1K