# What are the maximum and minimum values that the function f(x)=x/(1 + x^2)?

##### 1 Answer
Jun 21, 2015

Maximum: $\frac{1}{2}$
Minimum: $- \frac{1}{2}$

#### Explanation:

An alternative approach is to rearrange the function into a quadratic equation. Like this:

$f \left(x\right) = \frac{x}{1 + {x}^{2}} \rightarrow f \left(x\right) {x}^{2} + f \left(x\right) = x \rightarrow f \left(x\right) {x}^{2} - x + f \left(x\right) = 0$

Let $f \left(x\right) = c \text{ }$ to make it look neater :-)

$\implies c {x}^{2} - x + c = 0$

Recall that for all real roots of this equation the discriminant is positive or zero

So we have, ${\left(- 1\right)}^{2} - 4 \left(c\right) \left(c\right) \ge 0 \text{ "=>4c^2-1<=0" } \implies \left(2 c - 1\right) \left(2 c + 1\right) \le 0$

It is easy to recognise that $- \frac{1}{2} \le c \le \frac{1}{2}$

Hence, $- \frac{1}{2} \le f \left(x\right) \le \frac{1}{2}$

This shows that the maximum is $f \left(x\right) = \frac{1}{2}$ and the minimum is $f \left(x\right) = \frac{1}{2}$