find the two points on the curve

y=x^{4}-2x^{2}-x that have a common tangent line

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- Oct 10th 2012, 03:25 PMpnfullercommon tangent line?
find the two points on the curve

y=x^{4}-2x^{2}-x that have a common tangent line - Oct 10th 2012, 04:51 PMMaxJasperRe: common tangent line?
Try the two points:

x1=-1, x2=1

http://mathhelpforum.com/attachment....1&d=1349916630 - Oct 10th 2012, 04:54 PMpnfullerRe: common tangent line?
- Oct 10th 2012, 06:33 PMMarkFLRe: common tangent line?
The equation of the line to a function at $\displaystyle (x_n,f(x_n)$ is given in slope intercept form as:

$\displaystyle y=f'(x_n)x+f(x_n)-x_nf(x_n)$

Two lines are the same if they have the same slope and*y*-intercept. Using the points $\displaystyle (x_1,f(x_1))$ and $\displaystyle (x_2,f(x_2))$ will give you two equations and two unknowns, from which you can find two distinct points. - Oct 10th 2012, 07:01 PMx3bnmRe: common tangent line?
The slope of the tangent line is:

$\displaystyle m = \frac{dy}{dx} = 4x^3-4x-1$

$\displaystyle f'(a) = 4 a^3 -4a -1$

$\displaystyle f'(b) = 4 b^3 -4b -1$

$\displaystyle f(a) = a^4 -2a^2 -a$

$\displaystyle f(b) = b^4 -2 b^2 -b$

Now:

$\displaystyle f'(a) =\frac{f(b) - f(a)}{b -a}$

And:

$\displaystyle f'(b) =\frac{f(b) - f(a)}{b -a}$

So now the equation becomes:

$\displaystyle (4 a^3 -4a -1)(b - a) = (b^4 -2 b^2 -b) - (a^4 -2a^2 -a)..................(1)$

$\displaystyle (4 b^3 -4b -1)(b - a) = (b^4 -2 b^2 -b) - (a^4 -2a^2 -a)..................(2)$

Solving these two equations by maple we find $\displaystyle a = -1 \text{ and } b = 1$ or $\displaystyle a = 1 \text{ and } b = -1$ - Oct 11th 2012, 08:43 AMpnfullerRe: common tangent line?
- Oct 11th 2012, 09:09 PMhollywoodRe: common tangent line?
Maple is a program that does mathematical calculations. Here's a link:

Maple 16 by Maplesoft

- Hollywood