At what values of $x$ do $e^{3x}$ and $x^3$ have parallel tangent lines?

I was given this problem:

Let $$f$$ be the function defined by $$f(x)=e^{3x}$$, and let $$g$$ be the function defined by $$g(x)=x^3$$. At what value of $$x$$ do the graphs of $$f$$ and $$g$$ have parallel tangent lines?

To solve this problem, I set the derivatives of $$f$$ and $$g$$ equal to each other: $$3x^2=3e^{3x}$$. Now, I just need to solve for $$x$$, but for some reason I am completely at a loss of how to do that. I divided the problem by $$3$$, which got me $$x^2=e^{3x}$$. But, now what?

I also plugged it into my calculators solver, which got me the correct answer: $$-.484$$. But, I don't know how to find that answer without solver.

This problem can be solved using the Lambert-W function as follows. $$x^2=e^{3x}$$ $$\sqrt{x^2}=\sqrt{e^{3x}}$$ $$x=\pm e^{3x/2}$$ $$-\frac32 x=\mp\frac32 e^{3x/2}$$ $$-\frac32 xe^{-3x/2}=\mp\frac32$$ $$-\frac32 x=W_k\left(\mp\frac32\right)$$ $$x=-\frac23W_k\left(\pm\frac32\right)$$ for any branch of the function $$k\in\mathbb{Z}$$. The only real solution would be when $$k=0$$ and $$+$$ is taken for the $$\pm$$ giving $$x=-\frac23W_0\left(\frac32\right)\approx-0.4839075718\dots$$ which is the value provided by your calculator.

Since the other answer provides the exact solution using the Lambert $$W$$ function, let me show you how you can prove that the equation $$x^2 = e^{3x}$$ has one and only one solution using elementary methods.

Let $$F(x) = x^2$$ and $$G(x) = e^{3x}$$.

If you plot $$F$$ and $$G$$ on the same plane, you can see that there is exactly one point $$x$$ such that $$F(x) = G(x)$$. Here $$y = F(x)$$ is in red and $$y = G(x)$$ is in blue: Now, if you want to prove that there is a unique solution to the equation, you can reason as follows.

Consider first the interval $$(-\infty, 0]$$.

1. Since $$F$$ is decreasing and $$G$$ is increasing, there is at most one $$x \in (-\infty, 0]$$ such that $$F(x) = G(x)$$.
2. Since $$F(-1) > G(-1)$$ and $$F(0) < G(0)$$, there is at least one $$x \in (-1, 0)$$ such that $$F(x) = G(x)$$.

Combining the two results, we know that there is exactly one $$x \in (-\infty, 0]$$ such that $$F(x) = G(x)$$. Moreover, $$x \in (-1, 0)$$.

For the interval $$(0, \infty)$$ the situation is a bit trickier, because both functions are increasing. But if you know that $$x < e^x$$ for any $$x > 0$$, you can prove that $$F(x) < G(x)$$ as follows: $$x^2 < (e^x)^2 = e^{2x} < e^{3x}$$ Therefore there are no solutions in $$(0, \infty)$$.

• So would I be able to find the answer on a calculator by graphing the two equations and finding the intersection? – Burt Jul 31 at 17:36
• Yes. If you don't want (or aren't allowed) to use a graphing calculator, you can still plot the two graphs by hand. For a better approximation of $x$, you just need to find a sufficiently small interval $(a, b)$ such that $F(a) > G(a)$ and $F(b) < G(b)$, e.g. $(-0.49, -0.48)$. – Luca Bressan Jul 31 at 17:42

You can use Fixed-point iteration method to compute the root up to desired accuracy, which goes like this

$$x^2=e^{3x}\implies x=-e^{3x/2}=\phi (x)$$.....(1)

As $$\phi(x)$$ is continuous in $$[-1,0]$$ with $$\phi(x)\in [-1,0]$$ and the condition

|$$\phi'(x)$$|$$<1$$ is satisfied in $$[-1,0]$$, hence the iterative scheme from (1)

$$x_{n+1}=-e^{3x_n/2}$$ , $$n=0,1,2,...$$

is guaranteed to converge to the root $$-0.4839...$$ starting with any choice of $$x_0\in[-1,0]$$.