Question
At what point on the given curve is the tangent line parallel to the line 3x-y=6?
y=2+2e^(x)-3x
y=2+2e^(x)-3x
Answers
Reiny
The slope of the given line 3x - y = 6 is 3
dy/dx = 0 + 2e^x - 3
but that is equal to 3
2e^x - 3 = 3
2e^x = 6
e^x = 3
x = ln3
sub that into the function
y = 2 + 2e^ln3 - 3ln3
= 2 + 2(3) - 3ln3
= 8 + 3ln3
so it touches at the point (ln3,8+3ln3)
dy/dx = 0 + 2e^x - 3
but that is equal to 3
2e^x - 3 = 3
2e^x = 6
e^x = 3
x = ln3
sub that into the function
y = 2 + 2e^ln3 - 3ln3
= 2 + 2(3) - 3ln3
= 8 + 3ln3
so it touches at the point (ln3,8+3ln3)