Asked by berk
Use the instantaneous rate of change of f(x) = e^(5x) to find the equation of the tangent line to f(x) at x = 0.
Answers
Answered by
Steve
f(0)=1
f'(x)=5e^(5x)
So, f'(0)=5
Now we have a point and a slope, so the line is
y-1 = 5(x-0)
or
y=5x+1
See the graphs at
http://www.wolframalpha.com/input/?i=plot+y%3De%5E(5x),y%3D5x%2B1
f'(x)=5e^(5x)
So, f'(0)=5
Now we have a point and a slope, so the line is
y-1 = 5(x-0)
or
y=5x+1
See the graphs at
http://www.wolframalpha.com/input/?i=plot+y%3De%5E(5x),y%3D5x%2B1
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