Finding local max and min of a transcendental function

Question

So.. there's no way in heck i'll be solving the derivative of this algebraically -- what can I do? here's a hint from my professor..

Since the picture may be unclear, the function is $$f(x) = e^{3x} \log(x)$$ on $(0,\infty)$

this $\log(x)$ means NATURAL log here

"maybe it becomes easier if we have increasing and decreasing and 2nd derivative test criterions. you need critical points for 2nd derivative test. of course you wont be able to get explicit numbers, since you would have to solve transcendental equations. But you may be able to prove that there are say 2 critical points and determine which is the max and min?"

Answer 1

$\frac{d}{dx}[e^{3x} \log(x)]$

let $f(x) = e^{3x}$, $g(x) = \log(x)$

$\frac{d}{dx}[f(x)g(x)] = f'(x)g(x) + f(x)g'(x)$

$$\frac{d}{dx}[e^{3x}] \log(x) + e^{3x} \frac{d}{dx}[\log(x)]$$ $$(1)$$

Remember:

$\frac{d}{dx} [\log(x)] = \frac{d}{dx}[\frac{\ln x}{\ln 10}] = \frac{1}{\ln 10}\frac{d}{dx} [\ln x] = \frac{1}{x \ln 10}$

Also:

$\frac{d}{dx}[f(g(x))]=f'(g(x)) g'(x)$

so if $f(x) = e^x$ and $g(x) = 3x$

$\frac{d}{dx}[e^{3x}]=3e^{3x}$

Taking that other equation into consideration,

$$3e^{3x} \log(x) + \frac{e^{3x}}{x \ln 10} $$ is our final answer.

$3e^{3x} \log(x) + \frac{e^{3x}}{x \ln 10} = 0$

$3e^{3x} \log(x) = \frac{-e^{3x}}{x \ln 10}$. Dividing both sides by $e^{3x}$ gives us

$3 \log(x) = \frac{-1}{x \ln 10}$

$3x \log(x) = \frac{-1}{\ln 10}$

$3x \frac{\ln x}{\ln 10} = \frac{-1}{\ln 10}$

$3x \ln x = -1$

$\ln x e^{\ln x} = \frac{-1}{3}$

We can now use something called a Lambert W. Function. For any $xe^x = y$, $W(y) = x$

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

By raising both sides to the power of e, we get $e^{W(\frac{-1}{3})} = x$. Find a calculator or something online that evaluates the Lambert W. Function and use that to get your solutions for the first derivative, or critical points.

Answer 2

This answer is mainly intended to supplement the answer that louie mcconnell gave. Thus, I will not dwell on the mechanics of computing and rewriting the derivative, and instead I will focus on what seems to me to be the more difficult parts of the problem.

Answer Summary: There is a local maximum between $e^{-10}$ and $e^{-1},$ a local minimum between $e^{-1}$ and $e,$ and no other local extrema.

The derivative of $$e^{3x}\log x \; = \; e^{3x} \cdot \frac{\ln x}{\ln 10} \; = \; \left( \frac{1}{\ln 10} \right) \cdot \frac{e^{3x}}{\ln x}$$ is equal to

$$\left( \frac{1}{\ln 10} \right) e^{3x} \left(3 \, {\ln x} + \frac{1}{x} \right) $$ Since $\frac{1}{\ln 10}$ is positive and $e^{3x}$ is (always) positive, the derivative will be negative, zero, or positive according as $3 \, {\ln x} + \frac{1}{x}$ is negative, zero, or positive. Thus, if we use the first derivative test, we can replace the actual derivative with $\;3 \, {\ln x} + \frac{1}{x}.$

I think I know what your teacher has in mind in using the 2nd derivative test, but I think for this problem it is much easier to use the 1st derivative test. [For those interested, the 2nd derivative is a positive quantity times $9\,{\ln x} + \frac{6}{x} - \frac{1}{x^2}.$ If $\beta$ is a critical point of the original function, then $3\,{\ln \beta} = -\frac{1}{\beta},$ and so the 2nd derivative will be a positive quantity times $3\beta – 1.$ But there still remains the task of determining which critical numbers are greater than, less than, or equal to $\frac{1}{3}.]$

To find the critical points, we solve the equation $3 \, {\ln x} + \frac{1}{x} = 0.$ To me this seems by far the most difficult part of the problem. I will use the intermediate value property to show there are at least two solutions to this equation. If $x = e^{-10},$ then $3 \, {\ln x} + \frac{1}{x} = -30 + e^{10}$ is clearly positive. (I'm assuming we can use the fact that $e > 2.)$ If $x = e^{-1},$ then $3 \, {\ln x} + \frac{1}{x} = -3 + e$ is clearly negative. (I'm assuming we can use the fact that $e < 3.)$ If $x = e,$ then $3 \, {\ln x} + \frac{1}{x} = 3 + \frac{1}{e}$ is clearly positive. Therefore, there is a solution between $e^{-10}$ and $e^{-1},$ and there is a solution between $e^{-1}$ and $e.$ [Incidentally, what made me think of trying these values is they allowed me to get rid of the logarithm operation, and I was hoping (correctly, as it turned out) that when the logarithms are gone I might have a better chance of recognizing whether the values are positive or negative.]

However, it is not immediately clear whether there can be other solutions. To show there are no other solutions, I'll generalize what I did in the previous paragraph. Instead of more or less randomly picking values such as $e^{-10},$ $e^{-1},$ $e,$ let's see what happens if we pick an arbitrary number of the form $e^c,$ where $c$ is some real number. Note that the domain of the original function is all positive real numbers, and any positive real number can be written in the form $e^c$ for some real number $c.$ [Why? Let $r$ be a positive real number. Then picking $c = {\ln r},$ we have $r = e^{c}.]$ Therefore, solving the equation when $x = e^c$ will produce all the solutions (and no extraneous solutions) to the original equation.

Plugging $x = e^c$ into $3 \, {\ln x} + \frac{1}{x} = 0$ gives $3c + e^{-c} = 0,$ or $e^{-c} = -3c.$ To investigate the nature of the solutions, draw rough sketches by hand of $y = e^{-x}$ and $y = -3x$ together. (Any calculus student should be able to do this based on shifting/reflecting of "common graphs.) It should be clear that there are only two points of intersection, the left most intersection arising from the fact that $e^{-x}$ rises very rapidly as you move left through negative values of $x.$

Therefore, the two solutions to $e^{-c} = -3c$ give rise to two solutions to $3 \, {\ln x} + \frac{1}{x} = 0$ (via $x = e^{c}).$ Call these two solutions $x_1$ and $x_{2},$ where I've named them so that $x_1 < x_{2}.$ We know that $e^{-10} < x_1 < e^{-1}$ and $e^{-1} < x_2 < e$ from what I did above using the intermediate value property. In fact, since we know the derivative is positive at $x = e^{-10},$ negative at $x = e^{-1},$ and positive at $x = e,$ it follows from the first derivative test that $x = x_1$ corresponds to a local maximum and $x = x_2$ corresponds to a local minimum.