Re: how about this...

Posted by Soroban on August 20, 2002 at 00:20:55:

In Reply to: how about this... posted by T.Gracken on August 19, 2002 at 11:54:10:

: O.K. to write y= lx^2-3x+2l + lx^2-5x-14l - 3x + 1 without absolute values,

: start by determining when each absolute value expression is positive, negative, or zero.

: for lx^2-3x+2l, you should get x^2-3x+2 is positive on (-inf,1) and (2,inf), negative on (1,2)

: for lx^2-5x-14l, you should get x^2-5x-14 is positive on (-inf,-2) and (7,inf), negative on (-2,7)

: so consider the following intervals: (-inf,-2), (-2,1), (1,2), (2,7), and (7,inf).

: using the definition of absolute value,

: when x^2-3x+2 is positive, lx^2-3x+2l = x^2-3x+2,

: when x^2-3x+2 is negative lx^2-3x+2l = -x^2+3x-2,

: when x^2-5x-14 is positive, lx^2-5x-14l = x^2-5x-14, and

: when x^2-5x-14 is negative lx^2-5x-14l = -x^2+5x+14.

: {(because of html) i will use =< to mean less than or equal to.}

: using the intervals above, you can rewrite the function as

: y =

: 2x² - 11x - 11 ...if x =< -2

: -x + 17 ................................if -2 < x =< 1

: -2x² + 5x + 13 ...if 1 < x =< 2

: -x + 17 ................................if 2 < x =< 7

: 2x² - 11x - 11 ...if x > 7

:
: hope that helps.

In 38 years of teaching, I've never seen a problem
like that one - a great challenge.

I worked it out the same way: find the zeros of
the quadratics in absolute values, and considered
the function in each of those five intervals.

I was insufferably proud to have come up with the
same answers.

Well done, TG!

Follow Ups:

• Re: how about this... susan 10:10:33 08/20/02 (3)
  • your welcome and thanks to Mr."S" for checking my work & result. -nt- T.Gracken 13:20:14 08/20/02 (2)
    • that is supposed to be "you're welcome". T.Gracken 19:21:53 08/20/02 (1)
      • yeah, sure. fuck you T.Gracken. I see its ok to make mistatkes as long as your name is T.Gracken or Dr. Assdrew CoultAss. grumpy 18:58:47 08/22/02 (0)

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