Question

If
\[f(x)=\frac{ 2x^2-ax+1 }{ x^2+2x+2 }\]

Find all the real values of "a" that make, -1 12 years ago

Answer 1

Find the extrema of the function, then find \(a\) such that the extrema are no greater than 3 and no less than -1.

Answer 2

I totally agree with you @SithsAndGiggles but I seems to take alot of time doing that even using logarithimc diffirenation...

Answer 3

\[f'(x)=\frac{(4x-a)(x^2+2x+2)-(2x+2)(2x^2-ax+1)}{(x^2+2x+2)^2}\\
f'(x)=\frac{(4+a)x^2+6x-2(a+2)}{(x^2+2x+2)^2}\]
\(f'\) is 0 when
\[(4+a)x^2+6x-2(a+2)=0\\
x=\frac{-6\pm\sqrt{36-4(4+a)(-2(a+2))}}{2(4+a)}\\
x=\frac{-3\pm\sqrt{2a^2+12a+25}}{(4+a)}\]
So you have the restriction that \(a\not=-4\) and \(2a^2+12a+25\ge0\). You can ignore the second restriction, since it's true for all \(a\).

If \(x_1\) and \(x_2\) are the two critical points, the next step would be to find \(f(x_1)\) and \(f(x_2)\).

Answer 4

my friend @Juarismi I tell you what I think, it may be kinda confusing but I hope it's right, the graphs have shown me a satisfiyng results that's why I'm gonna post it.
okay we have that -1<f(x)<3 , we split the inequality into two inequalities so we get:
\[-1<f(x)\rightarrow -(x^2+2x+2)<x^2-ax+1\rightarrow 3x^2+(2-a)x+3>0\]
\[f(x)<3\rightarrow 2x^2-ax+2<x^2+2x+2 \rightarrow x^2+(6+a)x+5<0\]
I would like to notify you that the quantity \[x^2+2x+2=(x+1)^2+1\] is positive so I could multiply by it.
now I will take the first inequality for example:
we can solve for x to find that \[x=\frac{ (a-2)\pm \sqrt{(a-2)^{2}-36} }{ 6 }\] now to make the expression >0 set it as complete square ,thus\[(a-2)^2=36 \rightarrow a=-4,8\]
take a=-4 or 8. and you will find that the minimum value of f(x) is -1 but I will reject 8 because it gets the function out of the stated range. thus something>a>-4.
using the same logic for the second inequality you will find \[\sqrt{20}-6>a>something\]
so the final conclusion is:\[a \in (-4,\sqrt{20}-6)\]

Answer 5

dude @Juarismi do you know the final answer?, just to be sure of my solution..

Answer 6

Yes, your answer agree with the book.
It´s correct
Thanks.

Answer 7

okay now I can rest in piece :)
welcome any time dude