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OnMyWayToBeAMathProffesor · #12 January 1st, 2009, 04:01 PM

I see my mistake, so let me just show all my answer to makes sure the others are correct.

a) as $\lim_{x \to 0}f(x)$ $f(x)=\frac{1}{4}$ . I got this answer after i followed what Jhevon said and factored out certain terms.

b) $\frac{x^3-x}{x^3-4x}=\frac{x(x^2-1)}{x(x^2-4)}=\frac{x^2-1}{x^2-4}$ so $x^2-1=0$ thus $x=1$ , $x=-1$ correct?

c) $x^3-4x=0$ equals $x(x^2-4)=0$ thus zeros at $x=0$ $x=-2$ $x=2$ correct?

d)even because $f(x)=f(-x)$

thanks again for all your wonderful help.