To find the equations of the tangent lines to the curve y = x^3 + x + 1 that are parallel to the line given by 2x – y = 4, we can follow these steps:
Step 1: Find the slope of the given line
First, we need to determine the slope of the line. We can rearrange the equation of the line into slope-intercept form (y = mx + b):
y = 2x - 4
From this, we see that the slope (m) of the line is 2.
Step 2: Find the derivative of the curve
The next step is to find the derivative of the curve y = x^3 + x + 1 to get the slope of the tangent lines. The derivative (dy/dx) is:
dy/dx = 3x^2 + 1
Step 3: Set the derivative equal to the slope of the line
Now, we set the derivative equal to the slope we found earlier:
3x^2 + 1 = 2
Simplifying this gives:
3x^2 = 1
x^2 = 1/3
x = ±√(1/3)
Thus, the values of x where the slopes match are x = √(1/3) and x = -√(1/3).
Step 4: Find the corresponding y-values
Now, we need to find the corresponding y-values for these x-values using the original curve equation:
- For x = √(1/3):
y = (√(1/3))^3 + (√(1/3)) + 1
= (1/3√(1/3)) + (√(1/3)) + 1
= (1/3√(1/3)) + (3/3√(1/3)) + (3/3)
= (4/3√(1/3)) + 1
y = (-√(1/3))^3 + (-√(1/3)) + 1
= (-1/3√(1/3)) + (-√(1/3)) + 1
= (-1/3√(1/3)) - (3/3√(1/3)) + (3/3)
= (-4/3√(1/3)) + 1
Step 5: Find the equations of the tangent lines
With the x and y values, we can now use the point-slope form of the line equation y – y1 = m(x – x1) where m is the slope:
- For the point (√(1/3), (4/3√(1/3)) + 1):
y - ((4/3√(1/3)) + 1) = 2(x - √(1/3))
y - ((-4/3√(1/3)) + 1) = 2(x + √(1/3))
Final equations
Thus, the equations of the tangent lines to the curve that are parallel to the line 2x – y = 4 are:
- Tangent line at (√(1/3), (4/3√(1/3)) + 1)
- Tangent line at (-√(1/3), (-4/3√(1/3)) + 1)
These calculations give us the complete set of tangent lines that meet the specified condition of being parallel to the original line. Now you can plot these tangent lines along with the curve to visualize their relationship!