# What is the slope of the line normal to the tangent line of #f(x) = e^-x+x^2-x # at # x= 0 #?

Slope of the normal to

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To find the slope of the line normal to the tangent line of (f(x) = e^{-x} + x^2 - x) at (x = 0), we first need to find the derivative of (f(x)) and evaluate it at (x = 0) to find the slope of the tangent line. Then, the slope of the line normal to the tangent line will be the negative reciprocal of this slope.

The derivative of (f(x)) is (f'(x) = -e^{-x} + 2x - 1). Evaluating (f'(x)) at (x = 0), we get (f'(0) = -e^0 + 2(0) - 1 = -1).

So, the slope of the tangent line at (x = 0) is (-1).

The slope of the line normal to the tangent line will be the negative reciprocal of (-1), which is (1). Therefore, the slope of the line normal to the tangent line of (f(x)) at (x = 0) is (1).

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