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

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To find the equation of the line tangent to the function f(x)=(x^2-x)e^(1-x^2) at x=1, we need to find the derivative of the function at x=1 and then use the point-slope form of a line.

First, let's find the derivative of f(x): f'(x) = (2x-1)e^(1-x^2) - 2x(x^2-x)e^(1-x^2)

Next, substitute x=1 into the derivative to find the slope of the tangent line: f'(1) = (2(1)-1)e^(1-(1)^2) - 2(1)((1)^2-(1))e^(1-(1)^2)

Now, simplify the expression: f'(1) = (2-1)e^(1-1) - 2(1)(0)e^(1-1) f'(1) = 1

The slope of the tangent line is 1.

Now, we have the slope and a point (1, f(1)) on the tangent line. Let's find the y-coordinate of the point: f(1) = (1^2-1)e^(1-(1)^2) f(1) = 0

The point on the tangent line is (1, 0).

Using the point-slope form of a line, we can write the equation of the tangent line: y - 0 = 1(x - 1) y = x - 1

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