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Let [itex] f(x) = ax - \dfrac{x^3}{1+x^2} [/itex]

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:

[tex] 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} [/tex]

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

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:

[tex] 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} [/tex]

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

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