[B]Gratia Mindaro - Mathematics[/B]

Finally, it seemed as she had made it to the questions that were slightly more challenging than previous ones. She had been finding it utterly droll engaging in such repetitive pen strokes and finding no fulfilment in them. Gratia ... enjoyed the struggle. It was obvious that she would arise triumphant, but she hated a lack of challenge.

>If f(x) = (4x^4 - 12x^2)/3x then f'(x) equals ...

And .... it was another simple question. Why did she even bother?

f(x)	=(4/3)x^3 - 4x
f'(x)	=4x^2 - 4

>>Find the stationary points of f(x)

f'(x)	=4x^2 - 4
	=4(x^2 - 1)
	=4(x+1)(x-1)
0	=4(x+1)(x-1)
∴x=-1, x=1

f(-1)	=(4(-1)^4 - 12(-1)^2)/3(-1)
	=(4 - 12)/-3
	=-8/-3
	=8/3
∴(-1, 8/3)

f(1)	=(4(1)^4 - 12(1)^2)/3(1)
	=(4 - 12)/3
	=-8/3
∴(1, -8/3)

It would likely be a wise decision for her to progress onwards to chain, quotient and product rules.