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2005 - paper 1 - question 8 (C) - HSN forum

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2005 - paper 1 - question 8 (C)


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#1 xClairex

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Posted 14 May 2007 - 01:06 AM

i have managed to work out the maximum which = 9

not sure how to find the minimum value any help would be much appreciated

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#2 Steve

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Posted 14 May 2007 - 09:02 AM

This is an important point.

You have to remember that when a function has a restricted domain (-2\leq x\leq 2 in this case) then maxima and minima can occur at the endpoints of the interval.

This is much easier to see if you sketch the graph of the function in the question (which you can do since you have the turning points and the x- and y-axis intercepts).

If you do this you'll see a maximum turning point at x=0, giving the answer which you said.

As for the minimum, you're looking for where the graph has the smallest y-value it can but you're only allowed to have -2\leq x\leq 2. This boils down to you evaluating f(-2) and f(2). Which ever is the smallest is the minimum value (you can see this by your sketch - the only place it can be smallest is at one of the endpoints of the interval).

Hope this helps.
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#3 xClairex

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Posted 14 May 2007 - 10:38 AM

i understand what im doing now hopefully a question like this will turn up tomorrow

thank you






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