To find the equations of the tangent lines to the curve given by y = x³ – x + 1 that are parallel to the line x + 2y = 2, we need to determine the slope of the given line first.
We can rewrite the equation of the line in slope-intercept form (y = mx + b). Starting with:
x + 2y = 2
Subtracting x from both sides gives:
2y = -x + 2
Now, dividing everything by 2 yields:
y = -0.5x + 1
This tells us that the slope (m) of the line is -0.5.
Next, we need to find the slope of the tangent line at points on the curve, which requires us to compute the derivative of the curve:
y = x³ – x + 1
The derivative is:
dy/dx = 3x² – 1
To find the points where the tangent line is parallel to the given line, we set the derivative equal to the slope of the line:
3x² – 1 = -0.5
Solving this gives:
3x² = 0.5
x² = 0.5/3 = 1/6
x = ±√(1/6)
Now we find the corresponding y-values using the original curve equation:
For x = √(1/6):
y = (1/6)√(1/6) – (√(1/6)) + 1
And for x = -√(1/6):
y = – (1/6)√(1/6) + (√(1/6)) + 1
After evaluating both points, we can write the equations of the tangent lines in point-slope form. One of the equations will be:
y – y1 = -0.5(x – x1)
Where (x1, y1) is one of the points we found above. Repeat for the second point to get the second tangent line equation.
In conclusion, you will end up with two equations for the tangent lines that meet the requirements of the problem.