Curve Discussion with $ f(x) = \frac{1}{x^2+r} $ and $ r > 0 $ with r as a constant. Need guidance As stated, i have $$ f(x) = \frac{1}{x^2+r} $$
with r being a constant and $ r > 0 $
I am familiar with curve discussion normally, but confused by the constant r. How do i properly calculate this, without knowing the value of r, since r can be anything as long $ r > 0$? When i try to plot it i get some really crazy results.
Could anyone help me and show me for example how do i calculate properly the zero values, minima, maxima and so on? It would help a lot!
Thanks in advance everyone!
 A: To calculate the zeros, start by making the identity:
\begin{eqnarray}
f(x)=0=\frac{1}{x^{2}+r}.
\end{eqnarray}
Now you can easily see that the function gets to zero in the following limits:
\begin{eqnarray}
\lim_{x\to\pm \infty}f(x)=0,
\end{eqnarray}
as the denominator gets bigger and bigger for any finite $r>0$. Otherwise, there are no other values of $x$ that make the function identically zero.
Now coming to minima and maxima of any function $f(x)$, first you should calculate the first derivative of the function to identify its critical points, and equate it to zero:
\begin{eqnarray}
\frac{df(x)}{dx}=0.
\end{eqnarray}
There are several critical points where the first derivative can vanish. But only the values of $x$ where the derivative is equal to zero will be a critical point. Notice that the first derivative of a function is just telling you the gradient of that function at a concrete point. A maxima or minima must have a zero slope at the point where it occurs. But, another condition to determine whether the critical point is either a maxima or minima, is to take the second derivative:
\begin{eqnarray}
\frac{d^{2}f(x)}{dx^{2}}.
\end{eqnarray}
If the second derivative is $<0$, you have found a maxima of the function; if it is $>0$ then it is a minima, and if it is exactly $=0$ it is called an inflexion point. For the function above, try to make a plot and see where does a maxima/minima occur. The function you wrote is very standard.
