At what point on the curve
y = 6 + 2e^x − 3x
is the tangent line parallel to the line
3x − y = 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)
y = 6 + 2e^ln3 - 3ln3
= 6 + 2(3) - 3ln3
= 12 - 3ln3
To find the point on the curve where the tangent line is parallel to the given line, we need to find the point where the derivative of the curve is equal to the slope of the given line.
Here's how you can solve it step by step:
Step 1: Find the derivative of the curve
Differentiate the equation of the curve, y = 6 + 2e^x − 3x, with respect to x to find the derivative dy/dx.
dy/dx = 2e^x - 3
Step 2: Find the slope of the given line
Rearrange the equation of the given line, 3x - y = 1, into the slope-intercept form y = mx + c, where m is the slope.
Rewrite the equation: y = 3x - 1
Comparing it with the slope-intercept form, we can see that the slope (m) of the given line is 3.
Step 3: Equate the derivatives and slope
Set the derivative of the curve equal to the slope of the given line.
2e^x - 3 = 3
Step 4: Solve for x
Solve the equation to find the value(s) of x.
2e^x = 6
e^x = 3
Take the natural logarithm (ln) of both sides to solve for x:
x = ln(3)
Step 5: Find the corresponding y-coordinate
Substitute the value of x obtained in step 4 back into the equation of the curve to find the y-coordinate.
y = 6 + 2e^x − 3x
y = 6 + 2e^(ln(3)) − 3(ln(3))
Simplify:
y = 6 + 6 - 3ln(3)
y = 12 - 3ln(3)
Therefore, the point on the curve where the tangent is parallel to the line 3x - y = 1 is (ln(3), 12 - 3ln(3)).