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

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Answer 1 · 2015-12-06T04:26:50

#y=(4e^3-5)x-9e^3+9#

Finding #f'(3)# will tell us the slope of the tangent line when #x=3#. To figure out the point on the function, find #f(3)#.

#f(3)=3e^3-6#

The point will be #(3,3e^3-6)#.

Finding the derivative will require the product rule. Also remember that #d/dx[e^x]=e^x#.

#f'(x)=e^xd/dx[x]+xd/dx[e^x]-d/dx[x^2]+d/dx[x]#

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

#f'(3)=e^3+3e^3-6+1=color(brown)(4e^3-5#

Write in point-slope form:

#y-3e^3+6=(4e^3-5)(x-3)#

In slope-intercept form:

#y=(4e^3-5)x-9e^3+9#