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$f(x)=\frac{x^{2}-1}{x^{2}+1}$ for every real number for $x$, the minimum value of $f$ is what?
How can I find the minimum value of this function.I only know trial and error method, but it's not a generalized way.

Please tell me a generic way to solve this type of problem

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  • $\begingroup$ Differentiate it using the quotient and polynomial rules then set the derivative to zero. The x value will yield the minimum y. $\endgroup$
    – user117644
    Commented Aug 6, 2014 at 11:55
  • $\begingroup$ @mistermarko - you need to distinguish between maxima and minima, and also local maxima/minima and global ones. Since $\mathbb R$ is not compact, there is no guarantee that a maximum or minimum will be attained - as is the case with the maximum value here. $\endgroup$ Commented Aug 6, 2014 at 12:02

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This case is so simple, it can be solved without even using calculus. Write $f(x)$ as $$f(x)= 1- \frac{2}{x²+1} $$ The minimum value of this function clearly occurs when the fraction on the right is greatest, which clearly occurs when the value of the denominator is least, which happens when $x=0$ Hence, the minimum value of the function is $f(0)=-1$.

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    $\begingroup$ this is nice, but the op explicitly asked for a general method $\endgroup$
    – Ant
    Commented Aug 6, 2014 at 12:05
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    $\begingroup$ Yes, but the OP also stated that they only knew the trial and error method, so I assumed they had no knowledge of calculus. $\endgroup$ Commented Aug 6, 2014 at 12:06
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The domain of $f$ is $\mathbb{R}$

$f$ is differentiable.

$$f'(x)=\frac{2x(x^2+1)-(x^2-1)2x}{(x^2+1)^2}=\frac{4x}{(x^2+1)^2}$$

$$f'(x)=0 \Rightarrow x=0$$

$$f'(x)<0,\forall x<0$$

$$f'(x)>0, \forall x>0$$

Therefore, $f$ is decreasing on $(-\infty,0]$ and increasing on $[0,+\infty)$

So, $f$ achieves its minimum at $0$ and the minimum is equal to $f(0)=-1$

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    $\begingroup$ thanx it's very helpful for me $\endgroup$ Commented Aug 7, 2014 at 1:57
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Differentiating it gives you the answer :

$$f'(x)=\frac{2x(x^2+1)-(x^2-1)\cdot 2x}{(x^2+1)^2}=\frac{4x}{(x^2+1)^2}.$$ Since $f(x)$ is decreasing for $x\lt 0$ and is increasing for $x\gt 0$, $f(0)=-1$ is the min.

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For this particular $f(x)$ you can use polynomial division so that $x^2-1=1\cdot(x^2+1)-2$ as follows $$f(x)=\frac {x^2-1}{x^2+1}=\frac {x^2+1-2}{x^2+1}=1-\frac 2{x^2+1}$$

Since $x^2+1\ge 1$ it is then obvious that the minimum value occurs when $x=0$.


Added later: Note that polynomial division and also the use of partial fractions can help greatly in working out what is going on with rational functions like this.

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Others have covered how to solve the problem if $f$ is differentiable. But what if it's not?

Employ a computer and find a local minimum using the hill climbing algorithm. Or use a variant of it like "Random-restart hill climbing" for an approximation of the global minimum.

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