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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.

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Notice that the problem doesn't ask you to find the intersections, just conclude the number of them. Do you know how to graph the two curves? You can conclude the answer to this question with a thorough knowledge of the graphs. – Matthew Conroy Mar 6 '12 at 6:01

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 $x=2$.


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so? I am not getting "And y=x2+4x−5 is smaller than y=11+x2 when x=1, but is larger when x=2." – vikiiii Mar 6 '12 at 8:41
@vikiii: So if you look at the two graphs, one is going up, the other is going down. It would be possible that they never cross at all, if the one going down always stayed above the one going up (think of $y=1+\frac{1}{1+x^2}$, which never falls below $y=1$, and $y=1-\frac{1}{1+x^2}$ which never rises above $y=1$). But at $x=1$, $y=\frac{1}{1+x^2}$ has value $\frac{1}{2}$, while $y=x^2+4x-5$ has value $0$; and at $x=2$, $y=\frac{1}{1+x^2}$ has value $\frac{1}{5}$ and $y=x^2+4x-5$ has value $7$. So it cannot be the case that one always stays above the other. What does that tell you? – Arturo Magidin Mar 6 '12 at 15:20

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