Algebra step confusion and unnatural completing the square

[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

http://www.maths.cam.ac.uk/undergrad/admissions/step/advpcm.pdf

 

[jedishrfu]

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.

 

[Ray Vickson]

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.

 

[phospho]

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

 

[Karnage1993]

$f(x) = -f(-x)$

Multiply both sides by (-1)

$-f(x) = f(-x)$ or $f(-x) = -f(x)$

 

[phospho]

$f(x) = -f(-x)$

Multiply both sides by (-1)

$-f(x) = f(-x)$ or $f(-x) = -f(x)$

silly me, thanks

 

[HallsofIvy]

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

http://www.maths.cam.ac.uk/undergrad/admissions/step/advpcm.pdf

 

[phospho]

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.