# What is the equation of the normal line of #f(x)=(x-3)^3(x+2)# at #x=0#?

The equation of the normal line is

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To find the equation of the normal line of f(x)=(x-3)^3(x+2) at x=0, we need to find the slope of the tangent line at x=0 and then determine the negative reciprocal of that slope to find the slope of the normal line.

To find the slope of the tangent line at x=0, we take the derivative of f(x) and evaluate it at x=0.

The derivative of f(x) is f'(x) = 3(x-3)^2(x+2) + (x-3)^3.

Evaluating f'(x) at x=0, we get f'(0) = 3(0-3)^2(0+2) + (0-3)^3 = -27.

The slope of the tangent line at x=0 is -27.

The slope of the normal line is the negative reciprocal of the slope of the tangent line, so the slope of the normal line is 1/27.

Using the point-slope form of a line, we can write the equation of the normal line as y - f(0) = (1/27)(x - 0).

Simplifying, we get y - f(0) = (1/27)x.

Since x=0, we have y - f(0) = 0.

Therefore, the equation of the normal line of f(x)=(x-3)^3(x+2) at x=0 is y = f(0).

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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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