# A rudimentary differential calculus question

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

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? – Jp McCarthy Oct 25 '12 at 18:38

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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And you may want to treat the cases $a=0$ and $a \neq 0$ separately. – copper.hat Oct 25 '12 at 18:44
Well the answer I get when I follow your method is not correct.. – HarkamaljitSingh Oct 25 '12 at 18:48
@HarkamaljitSingh Well it's a very good approach, so it sounds like you made a mistake. Rather than dismissing the answer, you should include what you got or the work that you tried that led you to this conclusion. – rschwieb Oct 25 '12 at 19:16
Thanks, @copper.hat . I added a comment about that. – Matthew Conroy Oct 25 '12 at 20:15

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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The correction model says: $f'(0)=1.2$, so $a\geq1.2$ or $a\leq0$ – HarkamaljitSingh Oct 25 '12 at 18:55
I'm not quite sure what you mean by "correction model". Also, when looking for local extrema, you shouldn't evaluate the first derivative at 0, but rather set it equal to 0. For instance, the quantity $f'(0)$ didn't factor into my reasoning at all. – Alexander Sibelius Oct 25 '12 at 18:57
I meant the correction sheet I have, the 'answers'. Those are the correct answers, but the part $a\leq0$ canfuses me.. – HarkamaljitSingh Oct 25 '12 at 18:58
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. – Alexander Sibelius Oct 25 '12 at 19:05