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

By the Quotient rule we get

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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} ).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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