calculus help me

Question

Find a point on the graph of the function f(x)=3e^2x , where the tangent line to the graph is parallel to the line

y=3x-1

Answer 1

Since the requested tangent line must be parallel to y=3x-1 then its slope is m=3.

The tangent lines to f(x)=3e^2x are defined by its derivative f'(x).

We are interested in the point where f'(x) must equal 3.

The derivative of 3e^2x is 2(3e^2x)

So f'(x)=6e^2x =3

Or e^2x=3/6

Finding ln on both sides:

ln(e^2x )=ln(3/6)

2x = ln(0.5)

x= 0.5ln(0.5) and its corresponding y value is 3e^{2(0.5ln(0.5))} =1.5

The point exact coordinates are then (0.5ln(0.5) , 1.5)

The point approximate coordinates are then ( -0.347 , 1.5)