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?

1 Answer
Jun 17, 2018

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]}