$f(x)=\frac{x^{2}-1}{x^{2}+1}$ 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
$\endgroup$ 25 Answers
$\begingroup$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 $f(0)=-1$.
$\endgroup$ 2 $\begingroup$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 $f(0)=-1$
$\endgroup$ 1 $\begingroup$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.
$\endgroup$ $\begingroup$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.
$\endgroup$ $\begingroup$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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