# What is the equation of the line normal to #f(x)=2x^3 - x^2-x # at #x=4#?

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To find the equation of the line normal to the function f(x) = 2x^3 - x^2 - x at x = 4, we need to determine the slope of the tangent line at x = 4 and then find the negative reciprocal of that slope to obtain the slope of the normal line.

To find the slope of the tangent line, we take the derivative of the function f(x) with respect to x, which gives us f'(x) = 6x^2 - 2x - 1.

Substituting x = 4 into f'(x), we get f'(4) = 6(4)^2 - 2(4) - 1 = 96 - 8 - 1 = 87.

The slope of the tangent line at x = 4 is 87.

To find the slope of the normal line, we take the negative reciprocal of 87, which is -1/87.

Now, we have the slope of the normal line, and we also know that it passes through the point (4, f(4)).

Substituting x = 4 into f(x), we get f(4) = 2(4)^3 - (4)^2 - 4 = 128 - 16 - 4 = 108.

Therefore, the equation of the line normal to f(x) = 2x^3 - x^2 - x at x = 4 is y - 108 = (-1/87)(x - 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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