(x^3+e^2x+5)^(1/3) It just ends up oscillating between 0.854 & -2.884
Very good point! $\displaystyle a^{1/3}= 0$ if and only if a= 0. That will not have the vertical tangent at the solution point. Skeeter's point was that the derivative of $\displaystyle y= (x^3+ e^x+ 5)^{1/3}$, $\displaystyle y= (1/3)(x^3+ e^x+ 5)^{-2/3}(3x^2+ e^x)$ is infinite where $\displaystyle x^2+ e^x+ 5= 0$, exactly the point you are looking for.
If the equation to be solved is $\displaystyle x^{3} + e^{2x} + 5 =0$ the Newton-Raphson iterations are given by the difference equation...
$\displaystyle \displaystyle \Delta_{n} = x_{n+1} - x_{n} = - \frac{x^{3} + e^{2x} + 5}{3 x^{2} + 2 e^{2 x}} = \varphi (x_{n})$ (1)
The function $\displaystyle \varphi(x)$ is illustrated here...
There is only one 'attractive fixed point' in $\displaystyle x_{0} = -1.71366985564...$ and for any initial value the sequence (1) converges at $\displaystyle x_{0}$...
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$
It is interesting to observe that the search of the solution of the equation $\displaystyle \displaystyle (x^{3} + e^{2x} +5)^{\frac{1}{3}}$ with the Newton-Raphson method conducts to the difference equation...
$\displaystyle \displaystyle \Delta_{n} = x_{n+1} - x_{n} = -\frac{(x^{3} + e^{2x} +5)^{\frac{1}{3}}}{\frac{1}{3}\ {(x^{3} + e^{2x} +5)^{-\frac{2}{3}} (3 x^{2} + 2 e^{2 x})}} = - 3\ \frac{x^{3} + e^{2 x} + 5}{3 x^{2}+2 e^{2x}} = 3\ \varphi (x_{n})$ (1)
... where $\displaystyle \varphi(x)$ is the function described in my previous post. At first it seems that nothing changes but it isn't!... The function $\displaystyle 3\ \varphi(x)$ is illustrated here...
The 'attractive fixed point' is [of course...] again at $\displaystyle x_{0} = -1.71366985564...$ but [taht is the difference...] around $\displaystyle x=x_{0}$ is $\displaystyle |\varphi(x)|> |2\ (x_{0} - x)|$ [the 'red line' in figure...] and that means that the sequence obtained from (1) will never converge to $\displaystyle x_{0}$ ...
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$
The graph is positive-going.
When you pick an initial approximation to the root,
calculate the slope of the tangent,
find your next approximation and continue with the iterations,
your "next approximation" keeps going either side of the root.
There is one value of x that would cause you to "land on" the root.
But that would be pure luck!!