Finding the root using Newton's Method

54 Views Asked by At

The following exercise is in my study guide:

I need find the root of $f(x)=-e^{x}-4$, where $x_{0}=0$, performing at least 6 iterations.

Ok, I know that the derivative of $f(x)$ is

$\frac{\partial}{\partial x} \left ( -e^{x}-4 \right )=-e^{x}$

And I begin to perform the calculations of the iterations:

----Iteration #1----

$x_{1}=x_{0}-\frac{f\left ( x \right )}{{f}'\left ( x \right )}$

$x_{1}=0-\frac{-e^{x_{0}}-4}{-e^{x_{0}}}$

$x_{1}=-5$

----Iteration #2----

$x_{2}=x_{1}-\frac{-e^{x_{1}}-4}{-e^{x_{1}}}$

$x_{2} = −599.6526364$

----Iteration #3----

$x_{3} = x_{2} - \frac{-e^{x_{2}}-4}{-e^{x_{2}}}$

$x_{3} = −1.06632855\times 10^{261}$

----Iteration #4----

$x_{4} = x_{3} - \frac{-e^{x_{3}}-4}{-e^{x_{3}}}$

From here, the calculator throws me the following result

$x_{4} = undefined$

The graph of the function $f(x)=-e^{x}-4$ in GeoGebra it allowed me to observe that the function never touches the X axis. The only point I observed, was in (0, -5).

I do not understand what conclusion I can take out from the analysis of the iterations and the graph. Is it enough to say that the function $f(x)=-e^{x}-4$ does not have a root on the X axis? How could it be a better conclusion?