Answer:
Therefore, the minimum distance from the curve y = 2x^2 to the point (4, 0) is approximately 3.78 units, correct to 2 decimal places.
Explanation:
To find the minimum distance from the curve y = 2x^2 to the point (4, 0), we can follow these steps:
1. The distance between a point (x, y) on the curve and the point (4, 0) can be found using the distance formula:
- Distance = sqrt((x - 4)^2 + (y - 0)^2)
2. Substitute y = 2x^2 into the distance formula:
- Distance = sqrt((x - 4)^2 + (2x^2 - 0)^2)
3. Simplify the equation:
- Distance = sqrt((x - 4)^2 + 4x^4)
4. To find the minimum distance, we need to minimize the expression inside the square root. We can take the derivative of the expression with respect to x and set it equal to zero to find critical points.
5. Differentiate the expression:
- d/dx ((x - 4)^2 + 4x^4) = 2(x - 4) + 8x^3
6. Set the derivative equal to zero and solve for x:
- 2(x - 4) + 8x^3 = 0
- 2x - 8 + 8x^3 = 0
- 8x^3 + 2x - 8 = 0
7. Solve the equation for x using numerical methods or factoring. In this case, we can use numerical methods or a graphing calculator to find the approximate solution x ≈ 1.107.
8. Substitute the value of x back into the distance formula to find the minimum distance:
- Distance = sqrt((1.107 - 4)^2 + (2(1.107)^2 - 0)^2)
- Distance ≈ 3.78
Therefore, the minimum distance from the curve y = 2x^2 to the point (4, 0) is approximately 3.78 units, correct to 2 decimal places.