Thread: finding the turning points

Hi folks,

my mission is to find and determine the nature of the turning points on $y = e^x - ex^2$

so $\dfrac{dy}{dx} = e^x - 2ex = 0$ at a turning point

i.e. $x = \dfrac{e^x}{2e}$ ............................(1)

This is all very easy if you use wolfram or a graph plotter, but with only paper and pencil, not so easy!
I decided to use a graphical method to find the solutions of x for this equation.
That is for f(x) = 0, we rearrange to get g(x) = h(x) and plot y = g(x) and y = h(x) on the same diagram and find the intersection points. In this case $y = x$ and $y = \dfrac{e^x}{2e}$
BY plotting a few points, my diagram showed that I had roots in the area of 0.2 and 2.5
I now tried iterative techniques to get my values of x.

$x_{n+1} = \frac{1}{2e} e^x_{n}$

and with my initial value of $x_{0} = 0.2$ it converges to x = 0.232 in 5 cycles. But I cannot get the other root with this method!

I rearranged (1) to

$x_{n+1} = 1 + \ln 2 + \ln x_{n}$ and with $x_{0} = 2.5$ this iterates to my second root (2.678) in 8 cycles.

I could have used newton - raphson. The point is that in an exam, what do you do? It seems a bit of a lotto?

Any advice from the experts?

In an exam you shouldn't be given this sort of problem unless that exam is specifically on numerical methods of finding the zeros of a function, or at least that they had been recently covered.

If this course had nothing to do with this I think I'd leave my answer in terms of the simplest expression I could derive w/o using numerical methods.

In this case I'd simply say the turning points occur at the solutions of

$e^x = 2e x$

During such an exam I would probably try to confirm with the prof whether or not they wanted numerical methods to attempted to be used or not.

Thanks for your answer.
The question appeared at the end of a chapter on numerical methods.
The issue was that I wasn't sure which method to use. I wanted to verify that in some cases an iterative method will diverge, in which case you just try another until you get an answer where you expect it.
My book identifies the graphical, iterative and newton-raphson methods and it seems that any can work it just depends on the function which one actually does. That's fine. If one method always worked, there would only be one method!