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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: /0 on May 26, 2008, 12:30:39 am

Title: Stationary Points
Post by: /0 on May 26, 2008, 12:30:39 am: Find all stationary points to the equation $y=|x^{2m}-16x^{2m-2}|$ where m is a positive integer greater than or equal to 2.

I thought of trying to graph it then I got scared. I also tried it algebraically but the stationary points are mingled with the cusps so I can't sort them out. help thx
Title: Re: Stationary Points
Post by: enwiabe on May 26, 2008, 12:54:01 am: okay, for y = x^(2m) - 16x^(2m-2) > 0

x^2m(1 - 16x^(-2))> 0

thus, x^2m > 0 or (1 - 16/x^2) > 0

x^2m > 0 for all x, but 1 - 16/x^2 > 0 when x^2 - 16 > 0 or (x-4)(x+4) > 0 therefore, for all x > 4 and x <-4, (1 - 16/x^2) > 0.

So you're gonna have cusps at x = 4, and x = -4 ONLY.

I hope that helps you :)
Title: Re: Stationary Points
Post by: Mao on May 26, 2008, 03:57:09 pm: Quote from: DivideBy0 on May 26, 2008, 12:30:39 am
Find all stationary points to the equation $y=|x^{2m}-16x^{2m-2}|$ where m is a positive integer greater than or equal to 2.

I thought of trying to graph it then I got scared. I also tried it algebraically but the stationary points are mingled with the cusps so I can't sort them out. help thx

we can define this as a piecewise function [but its not necessary], but we first need to find its roots:

$0=x^{2m}-16x^{2m-2} = x^{2m-2} \cdot \left( x^2-16\right)$
$\implies x=-4,0,4$
and substituting gives us sign changes only at -4 and 4

$y=\left\{ \begin{array}{ccc} x^{2m}-16x^{2m-2} & x<-4 \\ 16x^{2m-2}-x^{2m} & -4 \le x < 4 \\ x^{2m}-16x^{2m-2} & x \ge 4 \\ \end{array}$

thus, cusps are at $x=\pm 4$ only.

differentiating the this function:

$\frac{dy}{dx}=\left\{ \begin{array}{ccc} 2mx^{2m-1}-16(2m-2)x^{2m-3} & x<-4 \\ 16(2m-2)x^{2m-2}-2mx^{2m-1} & -4 \le x < 4 \\ 2mx^{2m-1}-16(2m-2)x^{2m-3} & x \ge 4 \\ \end{array}$

$\frac{dy}{dx}=\left\{ \begin{array}{ccc} x^{2m-3}\cdot \left( 2mx^{2}-16(2m-2)\right) & x<-4 \\ x^{2m-3}\cdot \left( 16(2m-2)-2mx^{2}\right) & -4 \le x < 4 \\ x^{2m-3}\cdot \left( 2mx^{2}-16(2m-2)\right) & x \ge 4 \\ \end{array}$

$0=\left\{ \begin{array}{ccc} x^{2m-3}\cdot \left( 2mx^{2}-16(2m-2)\right) & x<-4 \\ x^{2m-3}\cdot \left( 16(2m-2)-2mx^{2}\right) & -4 \le x < 4 \\ x^{2m-3}\cdot \left( 2mx^{2}-16(2m-2)\right) & x \ge 4 \\ \end{array}$

$\implies \begin{array}{ccc} x^{2}=16-\frac{16}{m} & x<-4 \\ x=0\mbox{ or }x^{2}=16-\frac{16}{m} & -4 \le x < 4 \\ x^{2}=16-\frac{16}{m} & x \ge 4 \\ \end{array}$

$\implies x=0\mbox{ and }\pm 4\sqrt{\frac{m-1}{m}}$

alternatively, use the following derivative:

$\frac{d}{dx}|x| = sign(x), x\neq 0$

$\frac{dy}{dx} = \left(2mx^{2m-1}-16(2m-2)x^{2m-3}\right) sign\left(x^{2m}-16x^{2m-2}\right)$

this method, however, will not give you the repeated root [turning point] at x=0.
Title: Re: Stationary Points
Post by: /0 on May 26, 2008, 10:38:01 pm: Thanks so much!