How many times do these two graphs intersect for values x >0? When the curves
$$y = x^2 + 4x -5$$
and
$$y = \frac{1}{1+​x^2} $$
are drawn in the $xy$-plane, how many times do the two graphs intersect for values of $x > 0$ ?
I equate the value of $y$, then the equation comes in the fourth power of $x$ . How I can solve this?
Thanks in advance.
 A: While polynomials of degree 4 can be solved by radicals, that is not needed here.
The first graph, $y=x^2+4x-5 = (x+5)(x-1)$ is positive if $x\lt -5$ or if $x\gt 1$. It is increasing on $x\gt 1$. The graph of $y=\frac{1}{1+x^2}$ is always positive, and is decreasing on $x\gt 0$.
When  we look at the portions of the graphs that are on the first quadrant, the graph of $y=x^2+4x-5$ is going up, the graph of $y=\frac{1}{1+x^2}$ is going down.  And $y=x^2+4x-5$ is smaller than $y=\frac{1}{1+x^2}$ when $x=1$, but is larger when $x=2$.
So...
A: First, notice that these curves intersect in a point of absciss $x$
$$\Longleftrightarrow $x^2+4x-5 = \frac{1}{x^2 +1} \\
\Longleftrightarrow (x^2 + 1)(x^2+4x-5) = 1 \\
\Longleftrightarrow f(x) := x^4 + 4x^3 - 4x^2 + 4x - 6 = 0$$ 
For this problem you should study the variations of the function $f$. For this, look at the derivative of $f$, $f'(x) = 4x^3 + 12x^2 - 8x +4$. 
$f'(x)$ seems to be positive for $x \geqslant 0$, let's check this. For this, we will study the variations of $f'$. Yes, that means we'll have to take the derivative again, but no worry $f''(x) = 12x^2 + 24 x -8$ is a polynomial of degree 2.
The discriminant of $f''$ is $960$ and it has a negative root $x_0$ and a positive root $x_1 = \frac{-24 + \sqrt{960}}{24}$. So, we have
$$\begin{matrix}f''(x) \leqslant 0 & \textrm{if } 0\leqslant x \leqslant x_1 \\ f''(x) \geqslant 0 & \textrm{if } x \geqslant x_1 \end{matrix}$$
back to $f'$: Since $f'(x_1)\geqslant 0$, we get $f'(x)\geqslant 0$ for all $x\geqslant 0$.
back to $f$: $f$ is increasing on $\mathbb{R}^+$, and $f(0) = -6 < 0$. Furthermore, we have $\lim_{x\to +\infty} f(x) = +\infty$.
Hence, the equation $f(x) = 0$ has exactly one positive solution.
back to the question : The curves intersect only once for $x \geqslant 0$.
