# Thread: If the f(1)= 4

1. ## If the f(1)= 4

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

2. 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?

3. Originally Posted by wolfyparadise
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

Nor, do the rest of us, please post the entire question.

CB

4. sorry but this is the question which i posted is what i have , i also am having daoubt about the question thats why i posted it in the forum
please let em knw is any body can figure it out

It is a carelessly-worded problem.
As stated, there is an infinite number of possible answers.

If $f(1) = 4,\;f(2)=5, f(7) = 5,\;f(8)=4$
Find the value of $f(6).$
Also find the value of x for which $f(x)$ is maximum or minumum.
Graph the given points . . .

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

From the symmetry, I would guess that
. . we are expected to assume $f(x)$ is a parabola.

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

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

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

Now you can answer their question . . .