# What is the equation of the normal line of #f(x)= 1/x-e^(-x^3+x^2) # at #x=-1#?

The equation(point-slope form) of the normal line would thus be

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To find the equation of the normal line at x = -1, we need to determine the slope of the tangent line at that point and then find the negative reciprocal to obtain the slope of the normal line.

First, we find the derivative of f(x) using the quotient rule and the chain rule:

f'(x) = [(-1)(1/x^2) - (-e^(-x^3+x^2))(3x^2-2x)] / (x^2)

Next, we substitute x = -1 into f'(x) to find the slope of the tangent line at x = -1:

f'(-1) = [(-1)(1/(-1)^2) - (-e^(-(-1)^3+(-1)^2))(3(-1)^2-2(-1))] / ((-1)^2)

Finally, we take the negative reciprocal of f'(-1) to obtain the slope of the normal line:

m_normal = -1 / f'(-1)

This gives us the slope of the normal line.

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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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