My Math Forum real number x for which inequality

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November 25th, 2014, 04:48 PM   #1
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real number x for which inequality

Dear

need help solving the following question please

just one of the statements (a) – (d) is true

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 November 25th, 2014, 07:38 PM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,271 Thanks: 2435 Math Focus: Mainly analysis and algebra $$x \gt 3 \implies x^3 - 4x^2 - 7x + 12 \gt x - 3 \implies x^3 - 4x^2 - 8x + 15 \gt 0$$ This is clearly true for large positive $x$ so a) is false. For b), we look at $$x \lt 3 \implies x^3 - 4x^2 - 7x + 12 \gt 3 - x \implies x^3 - 4x^2 - 6x + 9 \gt 0$$ This is clearly false for large negative $x$, so b) is also false. The factor theorem tells us that $x=1$ is a root of $x^3 - 4x^2 - 6x + 9$. Now we note that the boundaries given in part c) fall either side of $x=1$ so the inequality can't be true for the whole range given. So c) is also false. So d) is true. I would argue that e) was also true because you posted the question here. Last edited by v8archie; November 25th, 2014 at 07:42 PM.
 November 25th, 2014, 08:00 PM #3 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,786 Thanks: 1029 Math Focus: Elementary mathematics and beyond $\displaystyle f(x)=x^3-4x^2-7x+12$ $\displaystyle g(x)=|x-3|$ a) False. f(x) is cubic and g(x) is linear so, at some point, f(x) is greater than g(x). b) False. If f(x) - g(x) = 0 has only one real root then there is only one (unbounded) interval over which the inequality holds. If f(x) - g(x) = 0 has three real roots then the inequality holds over one bounded interval and one unbounded interval, not two unbounded intervals. c) False. The solution to the inequality must include an unbounded interval. Hence choice d) is correct. Note: The domains of the unbounded intervals are specifically described as (left-bounded and right-unbounded)/left-open or (left-unbounded and right-bounded)/right-open.
 November 26th, 2014, 02:57 PM #4 Member   Joined: Nov 2014 From: london Posts: 32 Thanks: 1 thank u v8archie abut d answer i can say this only forum reply my question thank very much
 November 26th, 2014, 02:58 PM #5 Member   Joined: Nov 2014 From: london Posts: 32 Thanks: 1 thank u greg1313 thank u very much for ur time , Thanks from greg1313

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