In other words, when presented with the equation

$y=xe^x$

you can calculate the inverse function

$x=W(y)$

by utilizing its properties.

As an alternative solution, the Fixed-point iteration method can be employed to determine the root with the desired level of accuracy. This involves the use of the following equation:

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

…..(1)

Question:

I was given this problem:

Find the value of $x$ at which the graphs of $f(x)=e^{3x}$ and $g(x)=left(e^xright)^3$ have parallel lines.

In order to tackle the issue at hand, I equated the derivatives of two functions, namely $f$ and $g$. The resulting equation was $f”=3e^{3x}$. My next step is to find the value of $x$, but I’m struggling with this task for some reason. To simplify the problem, I divided it by $3$, which gave me $x^2=e^{3x}$. However, I’m unsure of how to proceed from here.

After connecting my calculators to the solver, I was able to obtain the accurate outcome of -$0.484. However, I am unaware of any other method to determine this solution without the use of the solver.

Solution 1:

To find a solution for the equation $x^2=e^{3x}$, one can use the Lambert-W function. The following steps can be taken: first, take the square root of both sides of the equation, giving $sqrt{x^2}=sqrt{e^{3x}}$. This can be simplified to $x=pm e^{3x/2}$. Multiplying both sides by $-frac32$ and switching the signs gives $-frac32 x=mpfrac32 e^{3x/2}$ and $-frac32 xe^{-3x/2}=mpfrac32$. Using the Lambert-W function, this can be further simplified to $-frac32 x=W_kleft(mpfrac32right)$, and finally $x=-frac23W_kleft(pmfrac32right)$ for any branch of the function, where $kinmathbb{Z}$. The only real solution occurs when $k=0$ and the positive sign is taken. This gives the value $x=-frac23W_0left(frac32right)approx-0.4839075718dots$, which matches the value provided by the calculator.

Solution 2:

As the previous response has already given the precise solution by utilizing the Lambert $W$ function, I will demonstrate an alternative elementary method to prove that the equation $x^2 = e^{3x}$ has a unique solution.

Define $F$ as a function of $x^2$, and let $G$ be the exponential function with base $e$ raised to the power of $3x$.

On a single plane, plotting both $F$ and $G$ reveals the existence of a solitary point $x$ where $F(x)$ equals $G(x)$. The red line represents $y = F(x)$, while the blue line represents $y = G(x)$.

To obtain the solution for the equation

prove that there is a unique

, you can use the following reasoning.

Initially, let’s examine the range of values within the interval denoted as $(-infty, 0]$.

- Given that $F$ is a decreasing function and $G$ is an increasing function, there exists at most one value of $x$ in the interval $(-infty, 0]$ such that $F(x) = G(x)$.
- Given that $F(-1) > G(-1)$ and $F(0) < G(0)$, there must exist a value of $x$ within the interval $(-1,0)$ where $F(x) = G(x)$.

From the combination of the two outcomes, it is evident that there is only a single value of $x$ in the range of $(-infty, 0]$ for which $F(x)$ is equal to $G(x)$. Additionally, it is worth noting that the value of $x$ lies within the interval of $(-1, 0)$.

The case for the interval $(0, infty)$ is more complex as both functions increase. However, knowing that $x

0$, it is possible to demonstrate that $F(x)

Solution 3:

One approach to finding the root of a function with a desired level of precision involves the use of the fixed-point iteration method. This method can be applied in the following manner.

The equation $x^2=e^{3x}$ can be solved by finding the root $x=-e^{3x/2}$, which can also be represented as $phi(x)$. This solution is given in equation (1).

Given that $phi(x)$ is a continuous function on the interval $[-1,0]$, such that its range is also contained within the same interval. Additionally, the following condition is satisfied:

The interval $[-1,0]$ satisfies the condition that $phi'(x) < 1$, as per the iterative scheme from (1).

For all non-negative integers $n$, the sequence $(x_n)$ is defined recursively by setting $x_0$ to be a constant and then setting $x_{n+1}$ to be equal to $-e^{3x_n/2}$.

No matter what initial value, within the range of $[-1,0]$, is selected, the root at $-0.4839…$ is assured to be reached by the method.