What are the local extrema, if any, of #f (x) =x/(-12x+2#?

By the Quotient rule we get

To find the local extrema of ( f(x) = \frac{x}{-12x + 2} ), we first find its critical points by setting the derivative equal to zero and solving for ( x ). Then, we determine whether these critical points correspond to local extrema.

Taking the derivative of ( f(x) ) with respect to ( x ), we get:

[ f'(x) = \frac{d}{dx} \left(\frac{x}{-12x + 2}\right) ] [ = \frac{d}{dx} \left(\frac{x}{-12x + 2}\right) ] [ = \frac{(1)(-12x + 2) - (x)(-12)}{(-12x + 2)^2} ] [ = \frac{-12x + 2 + 12x}{(-12x + 2)^2} ] [ = \frac{2}{(-12x + 2)^2} ]

Setting ( f'(x) ) equal to zero:

[ \frac{2}{(-12x + 2)^2} = 0 ]

This equation has no solution because a fraction can only equal zero if its numerator is zero, but the numerator is always 2, which is never zero. Therefore, there are no critical points and consequently no local extrema for ( f(x) = \frac{x}{-12x + 2} ).