Finding the max of $x^2/(2 + x^2)$ I tried to find the maximum value of the following function using the first derivative and equate it to zero :
The function :  $y=x^2 /(2+x^2)$
The first derivative: $4x/(x^2 +2)^2 =0 \implies x=0$ which gives an answer $y = 0$.
But this is not true. If we make $x = 1$, we get $y= 1/3$, which is bigger than 0!
Is there something I misunderstand 
 A: You found a zero of the first derivative. That does not mean that you found a maximum. It means you found a potential local minimum or local maximum.
In this case, you found the minimum. $0$ is the smallest the function will ever be. Note that this is easy to see, as everything is positive and/or squared, so you'll never get below $0$.
This function does not take a max. It gets arbitrarily close to $1$, but no better. If you think about what happens when $x \to \infty$, you'll see that this gets closer and closer to $1$. And since the denominator is always larger than the numerator, the functino will never actually reach $1$ $\diamondsuit$
A: the graph of $$y = \frac{x^2}{2+x^2}$$ is even has a global minimum zero at $x = 0.$ the graph is strictly increasings for $0 < x < \infty$ and approaches $y = 1.$  the function has no global maximum on $-\infty< x < \infty.$ the graph looks like a bell turned upside down.
A: You don't need any derivative to find the maximum of this function, just a small knowledge of inequalities: rewrite the function as:
$$f(x)=1-\frac 2{2+x^2}$$
The maximum of $f(x)$ corresponds to the minimum of $\dfrac 2{2+x^2}$, i.e. to the maximum of $2+x^2$. There is no such maximum, hence $f(x)$ has no maximum.
However, as $2+x^2\to+\infty$ as $x\to\pm\infty$, $f(x)$ has a least upper bound, which is equal to $1$.
Note: The first derivative test for a local maximum or a local minimum asserts that $f'(x)$ must be zero  at such a point. A sufficient condition for a local extremum is $f'(x)=0$  while changing sign at $x$.
On another hand, there may be a global extremum at a point where the derivative is not $0$, if it is not an interior point of the domain. It may also happen at a point where the function has no derivative.
A: Observe that
$$
\frac{x^2}{2+x^2}=1-\frac{2}{2+x^2}<1, \quad \forall x\in\mathbb{R},
$$
$$
\lim_{x\rightarrow\infty}\frac{x^2}{2+x^2}=1.
$$
Hence, we can not find $x_0\in\mathbb{R}$ such that 
$$
\frac{x_0^2}{2+x_0^2}=\max_{x\in\mathbb{R}}\frac{x^2}{2+x^2}.
$$
