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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

many thanks for your help
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November 25th, 2014, 07:38 PM   #2
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$$x \gt 3 \implies x^3 - 4x^2 - 7x + 12 \gt x - 3 \implies x^3 - 4x^2 - 8x + 15 \gt 0$$
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.
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November 25th, 2014, 08:00 PM   #3
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$\displaystyle f(x)=x^3-4x^2-7x+12$

$\displaystyle g(x)=|x-3|$

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.
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November 26th, 2014, 02:57 PM   #4
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Thumbs up thank u v8archie

abut d answer
i can say this only forum reply my question thank very much
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November 26th, 2014, 02:58 PM   #5
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Thumbs up thank u greg1313

thank u very much for ur time ,
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