Proving that function has only one root

Question

How to prove that the following function has only one root: $$f(x)= \frac{2x^2\ln(x)-x^2-8x}{4}$$ I tried with finding the second derivative and got $\ln(x)+1$. Now, since $\ln$ is an increasing function, the minimum is at $e^{-1}$. Then I tried to prove that the first derivative is always greater than zero by using $f'(x) \ge f'(\frac{1}{e})$ but on the right side I get a negative number. Is there another way to prove this without drawing graph.

Answer 1

Presumably you are only looking at $x>0$ (otherwise there might be argued to be a second root at $x=0$).

You might find it easier to consider $g(x)=\frac{4}{x^2}f(x)=2\ln(x)-1-\frac8x$ which has the same roots as $f(x)$ when $x>0$.

This $g(x)$ has derivative $g'(x)=\frac2x+\frac8{x^2}>0$ for $x>0$ so $g(x)$ is a continuous increasing function and so has at most one root.

Since $g(1)=2\times 0-1-8<0$ and $g(e^3)>2\times 3-1-\frac{8}{2^3}>0$, $g(x)$ must have exactly one root.

The root of both $f(x)$ and $g(x)$ is in fact $x = e^{W(4e^{-1/2}) + \frac12} \approx 4.23745494$, where $W(x)$ is the Lambert W function.

Answer 2

A simple approach (imo): your function is

$$f(x)=\frac12x^2\ln x-\frac14 x^2-2x\implies f'(x)=x\ln x-2<0\iff x<\alpha$$

where $\;\alpha\;$ is such that $\;\alpha\ln\alpha=2=\ln e^2\iff \alpha^\alpha=e^2\;$ (it's easy to see that $\;\alpha>1\;$ ...)

Thus your function begins decreasing monotonically up to $\;x=\alpha\;$ , and since

$$\lim_{x\to0^+} f(x)=0$$

we see that $\;f\;$ is negative on $\;(0,\,\alpha)\;$. For $\;x>\alpha\;$ the function increases monotonically (as its first derivative is then positive), and since

$$\lim_{x\to\infty}f(x)=\lim_{x\to\infty}\left(\frac{x^2}4\left[2\ln x-1\right]-2x\right)=\infty$$

and $\;f\;$ is continuous on $\;(0,\infty)\;$, it must cross the $\;x\,-$ axis exactly once as $\;f\;$ is injective (1-1) .