# Thread: finding the turning points

1. ## 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?

2. ## Re: finding the turning points

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.