the problem is f(x)=(x^3-3x)/(x^2-2)
I don't even know how to start I have to find the y and x intercepts first can someone tell me how i do that?
Then i tried to find the derivative but i don't think it was right because it's rly long.
PLEASE HELP
the problem is f(x)=(x^3-3x)/(x^2-2)
I don't even know how to start I have to find the y and x intercepts first can someone tell me how i do that?
Then i tried to find the derivative but i don't think it was right because it's rly long.
PLEASE HELP
No, oyu plug in y = 0 and x = 0 in the equation of the curve, not it's derivative.
And your derivative has a mistake.
$\displaystyle f'(x) = \dfrac{(x^2 - 2)(3x^2-3) - (x^3-3x)(2x)}{(x^2 - 2 )^2} = \dfrac{x^4 -3x^2 + 6}{(x^2 -2)^2}$
Oh, drat... This is a typo from me, sorry for that. I was pointing to the +6 which you put as -6 in your post.
$\displaystyle \begin{array}{cl}
f'(x) = \dfrac{(x^2 - 2)(3x^2-3) - (x^3-3x)(2x)}{(x^2 - 2 )^2} & = \dfrac{3x^4 - 3x^2 -6x^2 + 6 - (2x^4-6x^2)}{(x^2 - 2 )^2} \\
& = \dfrac{x^4 -3x^2 + 6}{(x^2 -2)^2}\end{array}$