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

2 Answers
Jul 8, 2016

I couldn't find any.


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

#f'(x) = 8e^x + 8xe^x - (2x-4)e^(x^2-4x)#

Critical point when #f'(x) = 0#

#therefore 8e^x + 8xe^x - (2x-4)e^(x^2-4x) = 0#

#e^x(8 + 8x) = e^(x^2-4x)(2x-4)#

Take natural logs of both sides:

#ln|e^x(8 + 8x)| = ln|e^(x^2-4x)(2x-4)|#

Using rules of logs we can rewrite as

#ln|e^x| + ln|8+8x| = ln|e^(x^2-4x)| + ln|2x-4|#

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

#x + ln|8+8x| = x^2 - 4x + ln|2x-4|#


#x^2 - 5x = ln|8+8x| - ln|2x-4| = ln|(8x+8)/(2x-4)|#

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

#e^(x^2-5x) = (4x+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(x)# given by:

#F(x) = e^(x^2-5x)(x-2) - 4x - 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 - (f(x_n))/(f'(x_n))#

and will have #x_(oo) = r#.

#F'(x) = e^(x^2-5 x) (x-2) (2 x-5)+e^(x^2-5 x)-4#


#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"# #color(blue)(x_0 = 5)#

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We end up with #r ~~ 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''(x) = e^(-4 x) (-4 e^(x^2) (x-2) x-2 e^(x^2)+8 e^(5 x)+40 e^(5 x) (x+1))-4 e^(-4 x) (8 e^(5 x) (x+1)-2 e^(x^2) (x-2))#

Subbing in our "r" value gives:

#f''(r) ~~ - 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.


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.