Intersection point

Problem

We are given the functions \(g:x \mapsto e^{-x}\) and \(h:x \mapsto x^3\) defined in \(\mathbb{R}\).

1. Illustrate by means of a sketch, that the graphs of \(g\) and \(h\) possess precisely one intersection.

2. Determine an approximate value \(x_1\) for the \(x\)-coordinate of this intersection by carrying out the first step of the Newton method for the function \(d:x \mapsto g(x)-h(x)\) defined in \(\mathbb{R}\) using as initial value \(x_0=1\).

Solution of part a

The graphs may easily be drawn by means of Sage:

Solution of part b

For an appropriately chosen initial value, the Newton method computes an approximate value for the zero nearby. In order to determine the zero of the function \(f(x)\), one needs to evaluate in each step

\[x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}\,.\]

For the first step one finds together with \(f(x)=d(x)=g(x)-h(x)\) and \(x_0=1\):

\[x_1 = 1 - \frac{g(1) - h(1)}{g'(1) - h'(1)} = 1 - \frac{e^{-1} - 1}{-e^{-1}-3}\approx 0{,}812\]

By means of Sage, several steps of the Newton method can be executed. As can be seen, already after five steps does the Newton method yield a result agreeing up to 12 digits with the zero determined numerically by Sage.