Find an equation for the line tangent to the graph of: f(x)=5xe^x at the point (a,f(a)) for a=2. what does y=?

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Answer 1 · 2018-06-27T01:58:58

#y=15e^2x-20e^2#

#(2,15e^2)#

Tangent Slopes are found with the derivative

#f'(x)=5xe^x+5e^x=5e^x(x+1)#

#f'(2)=15e^2#

#f(2)=10e^2#

Now we can construct the tangent line equation

#y=15e^2x-20e^2#

You're Y-Coordinate will just be the value of your original function at #x=2#

#(2,15e^2)#