Find the equation of the tangent line to the function at a given point f(x)=e^-3x+1 ; (0,e)?

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Answer 1 · 2018-01-21T10:30:02

#y-e=-3e(x-0)#.

From the given point I'm assuming the function is #f(x)=e^(-3x+1)# which contains the point #(0,e)#.

Using the Chain Rule on #f(x)# we have:

#f'(x) = e^(-3x+1)*(-3)#

So #f'(0) = -3e#.

The equation of the line can be written in point-slope form:

#y-e=-3e(x-0)#.

This can be rewritten as #y=-3e*x+e#