# What are the points of inflection, if any, of f(x)=8xe^x-e^(x^2-4x) ?

Jul 8, 2016

I couldn't find any.

#### Explanation:

We first find the critical points by evaluating the first derivative.

$f ' \left(x\right) = 8 {e}^{x} + 8 x {e}^{x} - \left(2 x - 4\right) {e}^{{x}^{2} - 4 x}$

Critical point when $f ' \left(x\right) = 0$

$\therefore 8 {e}^{x} + 8 x {e}^{x} - \left(2 x - 4\right) {e}^{{x}^{2} - 4 x} = 0$

${e}^{x} \left(8 + 8 x\right) = {e}^{{x}^{2} - 4 x} \left(2 x - 4\right)$

Take natural logs of both sides:

$\ln | {e}^{x} \left(8 + 8 x\right) | = \ln | {e}^{{x}^{2} - 4 x} \left(2 x - 4\right) |$

Using rules of logs we can rewrite as

$\ln | {e}^{x} | + \ln | 8 + 8 x | = \ln | {e}^{{x}^{2} - 4 x} | + \ln | 2 x - 4 |$

Because exp and ln are inverse operators, they cancel out:

$x + \ln | 8 + 8 x | = {x}^{2} - 4 x + \ln | 2 x - 4 |$

Rearranging:

${x}^{2} - 5 x = \ln | 8 + 8 x | - \ln | 2 x - 4 | = \ln | \frac{8 x + 8}{2 x - 4} |$

Take exponentials of both sides, we can also simplify the fraction on the right hand side:

${e}^{{x}^{2} - 5 x} = \frac{4 x + 4}{x - 2}$

This algebra isn't getting anywhere. We know that there is a critical point where f'(x) = 0. This point will be the root of this expression. We will have to find this root numerically. We define a new function $F \left(x\right)$ given by:

$F \left(x\right) = {e}^{{x}^{2} - 5 x} \left(x - 2\right) - 4 x - 4$

and will use the Newton-Raphson (NR) method to find the root of this function. NR is an iterative scheme based on an initial guess designed to find the zeroes of a function f such that f(r) = 0. It is given by

${x}_{n + 1} = {x}_{n} - \frac{f \left({x}_{n}\right)}{f ' \left({x}_{n}\right)}$

and will have ${x}_{\infty} = r$.

$F ' \left(x\right) = {e}^{{x}^{2} - 5 x} \left(x - 2\right) \left(2 x - 5\right) + {e}^{{x}^{2} - 5 x} - 4$

Hence

x_(n+1) = x_n - (e^(x_n^2-5x_n)(x_n-2) - 4x_n - 4)/(e^(x_n^2-5 x_n) (x_n-2) (2 x_n-5)+e^(x_n^2-5 x_n)-4

We will start with initial guess ${x}_{0} = 1$. I'll leave the number crunching out and just tabulate the results. color(red)("I started with "x_0 = 1 " and it took 127 iterations to stabilise"" Let's try that again with initial guess" $\textcolor{b l u e}{{x}_{0} = 5}$ We end up with $r \approx 5.376130311$. Nice, round number that. Anyway, this is the root we seem to end up at.

I'm going to shamelessly admit that I used Wolfram Alpha to do the second derivative of the original function because I'm lazy and it's just simple stuff, applied again and again.

$f ' ' \left(x\right) = {e}^{- 4 x} \left(- 4 {e}^{{x}^{2}} \left(x - 2\right) x - 2 {e}^{{x}^{2}} + 8 {e}^{5 x} + 40 {e}^{5 x} \left(x + 1\right)\right) - 4 {e}^{- 4 x} \left(8 {e}^{5 x} \left(x + 1\right) - 2 {e}^{{x}^{2}} \left(x - 2\right)\right)$

Subbing in our "r" value gives:

$f ' ' \left(r\right) \approx - 64968.9$ which is convincingly nowhere near zero. Therefore our root is not a point of inflection, it is a maximum.

I'm not getting any other roots despite using various other root finding methods, although looking at the plot there does seem to be a point of inflection. I can't find it so I'm going to tentatively say that there is no point of inflection, I'll happily eat my hat if someone can prove otherwise though!

Jul 9, 2016

I am just following Euan, in the search for point of inflexion , if any., See the explanation.

#### Explanation:

Please check whether there is a negative root for f''(x)=0, near 0. If so, Euan could use Newton Raphson method and approximate this root, for the point of inflexion x. It is relevant that f'(x) need not be 0, at a point of inflexion. For example, if f(x) = sin x, x=0 is a (tangent-crossing-the-curve) point of inflexion. f'(0)=cos 0 =1.