2
$\begingroup$

What is the minimum value of the $$ \frac {x^2 + x + 1 } {x^2 - x + 1 } \ ?$$

I have solved by equating it to m and then discriminant greater than or equal to zero and got the answer, but can algebraic manipulation is possible

$\endgroup$
  • $\begingroup$ You can find the minimum value of a function by setting the derivative of the function to zero and solving the resulting expression for $x$. $\endgroup$ – Daryl Jul 25 '12 at 4:42
  • $\begingroup$ Also, you can simplify slightly by noting that $\frac {x^2 + x + 1 } {x^2 - x + 1 } = 1+\frac {2x } {x^2 - x + 1 }$. $\endgroup$ – copper.hat Jul 25 '12 at 4:44
  • $\begingroup$ @Daryl: The question is labeled [algebra-precalculus], which would seem to mean we should check for non-calculus solutions. $\endgroup$ – Arturo Magidin Jul 25 '12 at 4:44
4
$\begingroup$

let y=$\frac{x^2+x+1}{x^2-x+1}$

=>$x^2(y-1)-x(y+1)+(y-1)=0$

As x is real, the discriminant= $(y+1)^2-4(y-1)^2≥0$

=>$(y-3)(y-\frac{1}{3})≤0$

=>$\frac{1}{3}≤y≤3$

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ +1 Cute technique; must have missed that with my 'new maths'... $\endgroup$ – copper.hat Jul 25 '12 at 6:23
  • $\begingroup$ To get from $1/3\leq y \leq 3$ to asserting that the minimum value is actually $1/3$, you need to demonstrate that $1/3$ is actually a possible value for $y$. In other words, you must show that there is a value of $x$ that gives this. Of course, there is, it's $x=-1$, but you really need to give this as part of your solution. $\endgroup$ – user22805 Jul 25 '12 at 9:06
  • $\begingroup$ +2n+1 :D This technique is so easy to understand and suitable <3 $\endgroup$ – William Phoenix Sep 5 '16 at 9:20
2
$\begingroup$

Here is an 'algebra solution':

$\frac {x^2 + x + 1 } {x^2 - x + 1 } = \frac{(x+1)^2-(x+1)+1}{(x+1)^2-3(x+1)+3} = \frac{1}{3} + \frac{2}{3} \frac{(x+1)^2}{x^2-x+1} = \frac{1}{3} + \frac{2}{3} \frac{(x+1)^2}{(x-\frac{1}{2})^2+\frac{3}{4}}$.

Since the last term is greater than zero when $x\neq -1$, we see that the minimum is

$\frac{1}{3}$.

Of course, this is cheating since I know the answer from J.D.'s solution and this suggests the way to expand the expression.

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

Hint: Take the derivative w.r.t $x$ and equate it with zero, you get: $$ \frac{d}{dx} \frac {x^2 + x + 1 } {x^2 - x + 1 } = - \frac {2(x^2 - 1)} {(x^2 - x + 1)^2} = 0.$$ So $x = \pm 1$ at the extrema. Test both for minimum.

Edit: this tutorial page might be helpful.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ But again, this is not algebra as possible. $\endgroup$ – user2468 Jul 25 '12 at 4:46
0
$\begingroup$

For $x\geq0$ we have $$\frac{x^2+x+1}{x^2-x+1}=1+\frac{2x}{x^2-x+1}\geq1.$$ For $x<0$ by AM-GM we obtain: $$\frac{x^2+x+1}{x^2-x+1}=1+\frac{2x}{x^2-x+1}=1+\frac{2}{x+\frac{1}{x}-1}=$$ $$=1-\frac{2}{-x+\frac{1}{-x}+1}\geq1-\frac{2}{2\sqrt{-x\cdot\frac{1}{-x}}+1}=\frac{1}{3}.$$ The equality occurs for $-x=\frac{1}{-x}$ or for $x=-1,$ which says that we got a minimal value.

By the same way we can get a maximal value if you want.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.