# generalized way of finding minimum value of a function?

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

• Differentiate it using the quotient and polynomial rules then set the derivative to zero. The x value will yield the minimum y.
– user117644
Aug 6 '14 at 11:55
• @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. Aug 6 '14 at 12:02

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 $f(0)=-1$

• thanx it's very helpful for me Aug 7 '14 at 1:57

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 $f(0)=-1$.

• this is nice, but the op explicitly asked for a general method
– Ant
Aug 6 '14 at 12:05
• Yes, but the OP also stated that they only knew the trial and error method, so I assumed they had no knowledge of calculus. Aug 6 '14 at 12:06

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.

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.

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.