# For #f(x)=1/x^3-1/(x-3)^3# what is the equation of the tangent line at #x=1/4#?

Equation of tangent is

The tangent appears as shown (not drawn to scale and shrunk vertically). graph{(1/x^3-1/(x-3)^3-y)(11243520x+14641y-3748608)=0 [-1, 1, 0, 150]}

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To find the equation of the tangent line at x=1/4, we need to find the derivative of f(x) and evaluate it at x=1/4. The derivative of f(x) is given by f'(x) = -3/x^4 + 3/(x-3)^4. Evaluating f'(x) at x=1/4, we get f'(1/4) = -3/(1/4)^4 + 3/(1/4-3)^4 = -768 + 3/(-47/4)^4 = -768 + 3/47^4/4^4 = -768 + 3/47^4/256. Therefore, the equation of the tangent line at x=1/4 is y = f(1/4) + f'(1/4)(x - 1/4).

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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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