Q) At what integral value of x will the function

(x^2 + 3x + 1)/(x^2 -3x+1) attains its maximum value?

Any help would be greatly appreciated.

Thanks,

Ashish

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- May 30th 2009, 02:41 AMa69356Maxima and Minima
Q) At what integral value of x will the function

(x^2 + 3x + 1)/(x^2 -3x+1) attains its maximum value?

Any help would be greatly appreciated.

Thanks,

Ashish - May 30th 2009, 03:36 AMfarlist
simplify the function->$\displaystyle 1+(6/(x-3+1/x))$

you will see that when x=1,it attains max and when x=-1,it attains min. - May 30th 2009, 03:47 AMCurious_eager
How do you simplify the above expression to 1+(6/(x-3+1/x)) ?

- May 30th 2009, 04:06 AMAmer
the critical points (3+5^1/2)/2 and (3-5^1/2)/2

since

(x^2 +3x+1)/(x^2 -3x+1)=(x^2-3x+1 +6x)/(x^2-3x+1) = 1+6x/(x^2-3x+1)

= 1+6(x-3+1/x)^-1= f(x)

f`(x) = -1 (1 -1/x^2)(x-3+1/x)^-2 = 0

$\displaystyle \frac{-1+\frac{1}{x^2}}{(x-3+\frac{1}{x})^2}$

the root of the numerator {-1,1} and the root denominator

$\displaystyle \frac{3-+\sqrt{5}}{2}$

but the denominator is always positive

look at this

Attachment 11682