$f(x) = e^{3x}\ln x$ has two local extrema

Question

I found the following task:

Show that $f(x) = e^{3x}\ln x$ has two local extrema on $]0,\infty[$.

The function $f(x)$ is monotonic increasing for $x>1 $ and $f(x) \to -\infty$ for $x\to +0$. Thus if there are any local extrema, then they must be on the interval $]0,1[$.

Thus I have to look for $f'(x)=0$ in $]0,1[$.

Now $f'(x) = e^{3x} ( 3\ln(x) + \frac{1}{x})$. Thus I need to find two points with

$$\ln(x) = -\frac{1}{3x}.$$

How does one solve this?

I thought maybe I could just insert random numbers from $]0,1[$ in the function I might be lucky to argue with continuity that such points exists. But I don't think that it is expected to insert just some random numbers to solve this?

Answer 1

What you did is correct. Now, consider the function $$ g(x)=\ln x+\frac{1}{3x}=\frac{3x\ln x+1}{3x} $$ and note that $\lim_{x\to0^+}g(x)=\infty$, $\lim_{x\to\infty}g(x)=\infty$.

Moreover, $$ g'(x)=\frac{1}{x}-\frac{1}{3x^2}=\frac{3x-1}{3x^2} $$ which vanishes for $x=1/3$, where $g$ has an absolute minimum. Since $$ g(1/3)=-\ln3+1<0 $$ we see that there are exactly two points where $g$ (and so $f'$) vanishes. These two points are indeed extremal points for $f$ (finish it up).

Answer 2

As other commenters have said, the equation $$\ln(x) = -\frac{1}{3x}$$ cannot be solved algebraically, but you don't need to solve it algebraically for this question. In particular, all you need to do is show that two solutions exist, not necessarily find them. One method for doing this might be the intermediate value theorem, that is, if $f'(a) < 0$ and $f'(b) > 0$, then you know there is some point $c$ in $]a, b[$ with $f'(c) = 0$.