Question

Find all points on the curve y = (17/2)x² - x³ where the tangent line is horizontal.

Answer 1

To find the points on the curve y = (17/2)x² - x³ with a horizontal tangent, calculate the first derivative, set it to zero, factor the resulting equation, solve for x, and then find the corresponding y values. The points with a horizontal tangent are (0, 0) and (17/3, 289/3).

Step-by-step explanation:

Finding Points with a Horizontal Tangent

To find all points on the curve y = (17/2)x² - x³ where the tangent line is horizontal, you need to determine where the derivative (slope of the tangent line) is equal to zero. A horizontal tangent line has a slope of zero, hence we need to find the derivative of the given function and set it to zero.

1. Find the derivative of the function: y' = d/dx[(17/2)x² - x³] = 17x - 3x².
2. Set the derivative equal to zero: 17x - 3x² = 0.
3. Factor the equation: x(17 - 3x) = 0.
4. Solve for x: x = 0 or x = 17/3.
5. Substitute x values back into the original equation to find y values: y(0) = 0 and y(17/3) = (17/2)(17/3)² - (17/3)³.
6. Calculate the y values: y(0) = 0 and y(17/3) = 289/3.
7. Therefore, the points on the curve with a horizontal tangent are (0, 0) and (17/3, 289/3).