If f(1) = 4 , f(2)=5, f(7) = 5 and f(8)=4, find the value of f(6).Also obtain the value of x for which f(x) is maximum or minumum.

I don't have a clue what to do

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- Jan 1st 2010, 10:42 PM #1

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- Jan 1st 2010, 11:00 PM #2

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Is this the complete problem? I mean f(6) can be anything really; I am also assuming you are meant to notice that f(x) is continuous, and thus a local maximum must occur between f(5) and f(7) - or hell f(5) and f(7) could be the local maximum and the graph dips btween f(5) and f(7). I dunno, theres a lot that can be going on here. Is this all the information - was there a picture to go along with this question?

- Jan 1st 2010, 11:50 PM #3

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- Jan 2nd 2010, 04:34 AM #4

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- Jan 2nd 2010, 08:27 AM #5

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Hello, wolfyparadise!

It is a carelessly-worded problem.

As stated, there is an infinite number of possible answers.

If $\displaystyle f(1) = 4,\;f(2)=5, f(7) = 5,\;f(8)=4$

Find the value of $\displaystyle f(6).$

Also find the value of x for which $\displaystyle f(x)$ is maximum or minumum.

Code:| 5 + * * | : : 4 + * : : * | : : : : | : : : : | : : : : . . + - + - + - - - - + - + - - - - - | 1 2 7 - 8 |

From the symmetry, I wouldthat*guess*

. . we are expected to assume $\displaystyle f(x)$ is a parabola.

The general parabola is: .$\displaystyle f(x) \:=\:ax^2 + bx + c$

And we can determine that: .$\displaystyle a = -\tfrac{1}{7},\;b = \tfrac{10}{7},\;c = \tfrac{19}{7}$

Hence, the function is: .$\displaystyle f(x) \;=\;-\tfrac{1}{7}x^2 + \tfrac{10}{7}x + \tfrac{19}{7}$

Now you can answer their question . . .