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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?

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Jun 18, 2018

Answer:

#y=5x+1#

Explanation:

The rate of change function of #f(x)# is its first derivative #f'(x)=5e^(5x)#. At #x=0#, #f'(0)=5#.

This is also the gradient of the line tangent to the curve, #y=mx+c#. So #m=5#.

To determine #c#, we make the values of the two curves match at the same point, #x=0#. #f(0)=1#, so #1=5*0+crArrc=1#.

Thus the equation of the tangent line at #x=0# is #y=5x+1#.

Sanity check the answer by plotting the two curves on the same graph and observing that they touch in the right place:
graph{(y-(e^(5x)))(y-(5x+1))=0 [-1, 1, -0.206, 2.291]}

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