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If f(x)=i, where i is an integer such that i<=x<i+1, and g(x)=f(x)-2x, then the period of g(x) is

(a) 1/2
(b) 1
(c) 2
(d) -2
(e) none of the above

<= means less than or equal to.

By Tweaks (Tweaks) on Saturday, May 24, 2003 - 04:46 pm: Edit

I remember that problem in the Barron's book.

E) None of the above.

There is no period, because there is no repetition of the same curve, like there exists in the sine and cosine graphs. Rather, g(x) has a -2x term which means that it keeps on falling.

By Peanuts (Peanuts) on Saturday, May 24, 2003 - 09:57 pm: Edit

what about this one:

|x-2|is less than or equal to 5.
how many integers ar eint he solution set?

the answer is 11, but i only count ten. and the stupid barrons offered some ridiculous explanation thats written in a different language.
i thought it was 10: 7,6,5,4,3,2,1,0,-1,-2. i think the barrons included 8 as a solution for some reason. 8-2=6, but 6 shouldnt work!

is it a typo or what?

By Peanuts (Peanuts) on Saturday, May 24, 2003 - 09:58 pm: Edit

NEVERMIND i counted wrong (i left out -3).
please disregard the above post on the count of my stupidity

By Andrew123s (Andrew123s) on Saturday, May 24, 2003 - 10:16 pm: Edit

what about (-1/16)^(2/3)?
is it .16 or imaginary? I get two different answer on the ti-89 depending on whether I'm in a+bi mode or real mode.

By Peanuts (Peanuts) on Saturday, May 24, 2003 - 11:08 pm: Edit

x^2+y^2=1 is an even function, but it's not an odd function right? my barron book says its both.

By Andrew123s (Andrew123s) on Saturday, May 24, 2003 - 11:43 pm: Edit

x^2+y^2=1 is not a function. It is both even and odd (because it is symmetric to the y axis and the origin), but it is not a function.

By Dschnapps (Dschnapps) on Sunday, May 25, 2003 - 02:57 pm: Edit

to andrew's question, it should be a real number, because: the two is on top, so you square it, and then cube root. You can cube root a negative number, so it shouldn't be imaginary.