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

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Answer 1 · 2017-02-01T04:15:24

The equation is #y = x#.

Start by finding the corresponding y-coordinate of tangency.

#f(1) = 3(1)^2 - 3(1) + e^(1 - 1^2) = 3 - 3 + e^0 = 1#

Differentiate now, using #d/dx(x^n) = nx^(n - 1)# and #d/dxe^(f(x)) = f'(x)e^(f(x))#.

#f'(x) = 6x - 3 - 2xe^(1 - x^2)#

The slope of the tangent is given by evaluating your point #x = a# into the derivative.

#f'(1) = 6(1) - 3 - 2(1)e^(1 - 1^2) = 6 - 3 - 2 = 1#

Find the equation of the tangent.

#y - y_1 = m(x-x_1)#

#y - 1 = 1(x - 1)#

#y - 1 = x - 1#

#y = x#

Hopefully this helps!