How do I prove that $f(x) = ln(x) − (x − 4)^2$ has exactly two roots.

Question

I assume we take the derivative of the function. I get: $y' = 1/x-2(x-4)$ and I attempt to set it to 0 and solve but get stuck. Any tips?

Answer 1

From $$ \frac1x=2x-8, $$ you multiply both sides by $x$ to get $$ 1=2x^2-8x, $$ of $x^2-8x-1$. Since $x^2-8x-1=(x-4)^2-17$, its roots are $4\pm\sqrt{17}$. But our function only exists for $x>0$, and so in its domain the derivative has a single root.

Now, $$ f(1)=-9,\ \ f(4)=\log4>0. $$ This guarantees a root for $f$ in $(0,4)$. We also have that $\lim_{x\to\infty}f(x)=-\infty$, so eventually $f$ has to be negative again, so it crosses the $x$ axis again: a second root. And there cannot be another one, because to "climb up" again and cross the $x$ axis one more time, the derivative would be zero at a second point.

Answer 2

Outline: 1) $f(1)\lt 0$, $f(4)\gt 0$, $f(10)\lt 0$, so by the Intermediate Value Theorem $f(x)=0$ has at least two roots. 2) By the Mean Value Theorem, between any two roots of $f(x)=0$ there is at least one root of $f'(x)=0$. 3) The equation $f'(x)=0$ has exactly one positive root.

Answer 3

Hint:

The roots are given by $$ \log x=(x-4)^2 $$

when we have so simple functions the first approach is to plot a graph, as in the figure, where we represents the two functions. Tis is simple because you know (I suppose) the graph of $y=\log x$ and the graph of $y=(x-4)^2$ is a positive parabola with vertex at $(4,0)$.(see the figure).

And this gives you a first intuition of the possible solutions that are the abscissa of the intersection points of the two graphs. Not a proof, but than you can refine, if you want, with more analytical considerations as in the other answers.