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What is the slope of the line normal to the tangent line of #f(x) = x^2e^(x-1)+2x-2 # at # x= 1 #?

Answer 1

#-1/4#

We know that a normal line is orthogonal to a tangent line, so we should find the slope of the tangent line first, then find the slope orthogonal to it.

The slope of the tangent line is just the derivative: #f'(x) = 2x e^(x-1) + x^2 e^(x-1) + 2 # #f'(1) = 2e^0 + 1e^0 + 2 = 4 #

Two orthogonal lines have #m_1m_2 = -1 #

so #m_(norm) * 4 = -1 implies m_(norm) = -1/4#

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Answer 2

Charles Arocha

To find the slope of the line normal to the tangent line of (f(x) = x^2e^{x-1} + 2x - 2) at (x = 1), first find the derivative of (f(x)), then find the slope of the tangent line at (x = 1), and finally determine the negative reciprocal of this slope to find the slope of the line normal to the tangent line.

(f'(x) = (2xe^{x-1} + x^2e^{x-1}) + 2)

(f'(1) = (2e^0 + e^0) + 2 = 3 + 2 = 5)

The slope of the tangent line at (x = 1) is (5).

The slope of the line normal to the tangent line at (x = 1) is the negative reciprocal of the slope of the tangent line, so it is (-\frac{1}{5}).

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