# Possible number of solutions (Rolle's and The Mean Value Theorems)

Help me please with this one...

What is possible number of solutions of equation: $$x^{2}-x- \ln x=a$$ for different values of $a$?

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Quick observation: You can see geometrically that $\ln(x)=x^2-x-a$ can have at most two solutions just by considering the shapes of the graphs, $\ln$ being concave down and $y=x^2-x-a$ being concave up. –  Jonas Meyer Jan 28 '12 at 20:41
@Jonas: Not that the amount of effort shown by the OP warrants further explanation, but you're definitely on a good train of thought. I'd upvote an answer that goes on to say when there are 0, 1 solutions. –  The Chaz 2.0 Jan 28 '12 at 21:00

## 1 Answer

Jonas' comment already settles the question of the maximum number of zeros that $f$ may have. I'll try to write down a more detailed proof.

Let $f : (0, +\infty) \rightarrow \mathbb{R}$ be the function $f(x) = x^2 - x -\ln(x) -a$, so that the solutions of your equation are precisely the zeros of $f$.

Now, suppose $f$ has two distinct zeros $x_1, x_2 \in (0, \infty)$. Then, applying Rolle's Theorem, you get that $f^{\prime}(\xi) = 0$ for some $\xi \in (x_1, x_2)$. The key observation (which is just a particular case of Rolle's Theorem) is this: between any two zeros of $f$, there must be a point where the derivative of $f$ is zero. If we know for some reason that the derivative of $f$ vanishes at most $n$ times, then $f$ has at most $n+1$ zeros.

In this case, $\xi$ must be positive, since $f(x)$ is only defined for positive $x$. But $f^{\prime}(\xi) = 0$ is just the equation $$2\xi -1 -\frac{1}{\xi} = 0,$$ as you can see directly by deriving $f$. The above equation has only one positive solution, namely, $\xi = 1$. This means that $f$ has at most two zeros. But this is just an upper bound; what else can we say about the exact number of zeros?

Since $f^{\prime\prime}(1) > 0$, the point $x = 1$ is a local minimum for $f$.

For $x > 1$, we have $f^{\prime}(x) > 0$, so that $f$ is increasing (to $+\infty$) after $1$.

For $x < 1$, we have $f^{\prime}(x) < 0$, so that in $(0, 1)$, the graph of $f$ decreases (from $+\infty$) to approach the value $f(1) = -a$.

The above analysis should convince you that the graph of $f$ around $x = 1$ looks pretty much like $x^2$ around $x = 0$. Finally, this leads to the conclusion that for $a = 0$, your original equation has only one solution ($x_1 = 1$); for $a > 0$ it has two solutions $x_1, x_2$ with $0 < x_1 < 1 < x_2$ and for $a < 0$ it has no solutions.

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