Question

At what point on the curve
y = 6 + 2e^x − 3x
is the tangent line parallel to the line
3x − y = 1?

Answer 1

dy/dx = 2e^x - 3

since tangent is to be parallel, its slope must be 3

2e^x - 3 = 3
2e^x = 6
e^x = 3
x = ln3

then y = 6 + 2e^ln3 - 3ln3
= 6 + 2ln3 - 3ln3 = 6 - ln3

the point is (ln3, 6-ln3)

Answer 2

y = 6 + 2e^ln3 - 3ln3
= 6 + 2(3) - 3ln3
= 12 - 3ln3