Originally Posted by
x3bnm 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 $\displaystyle a = 1 \text{ and } b = -1$