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$f(x)=\dfrac{6x}{x^2+5}$

Find out algebraically for which $a$ the function $f(x)=a$ has exactly 1 solution.

I haven't got a clue how to tackle this at all..

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    You should use algebra and the geometry of the graph of a quadratic. $f(x)=a$ $\Rightarrow ax^2-6x+5a=0$. What does this look like?2012-10-25

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What you want to do is look at solutions to the equation $ \frac{6x}{x^2+5} = a, $ that is, $ax^2-6x+5a=0.$

If $a=0$, this reduces to $-6x=0$. This clearly has one solution, $x=0$.

If $a \neq 0$, you can use the quadratic formula to solve for $x$ and see that the solutions depend on $a$, and, more specifically, the discriminant (the "$b^2-4ac$" part) depends on $a$, so the number of solutions depend on $a$.

There will be exactly one solution if and only if the discriminant is zero.

So, figure out what $a$ needs to be to make the discriminant equal to zero.

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    Thanks, @copper.hat . I added a comment about that.2012-10-25
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If you'd like to use differential calculus, try this method. We're dealing with a polynomial of degree 2. So, it will have at most two real roots. What does the graph look like if it has exactly one real root? In this case, it will have achieve a local extremum of $f(x)=0$.

The first derivative of $ax^2-6x+5a$ is $2ax-6$. Since the first derivative is linear, it will only have one root corresponding to one extremum. We find that this extremum occurs at $x=3/a$. Substitute this back into the original function. Setting this new polynomial in $a$ equal to $0$ and solving gives you the value of $a$ at which the function has a local extreme value of 0, which is what you want.

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    Well, when $a=0$, then we have the trivial linear polynomial $6x=0$ which clearly has one root. However, if $a$ equals, say, -1, then we have $two$ roots at $x=-1$ and $x=-5$. Therefore, the answers you were given are wrong. Both the answer I gave and the answer @Matthew Conroy gave are correct.2012-10-25