Find the coordinates of the point on the curve y = x^3 - 3x + 2 where the tangent is horizontal.

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Q: Find the coordinates of the point on the curve y = x^3 - 3x + 2 where the tangent is horizontal.

Solution: f'(x) = 3x^2 - 3. Setting f'(x) = 0 gives x = 1. The point is (1, 0).

Steps: 10

Step 1: Understand that we need to find where the tangent to the curve is horizontal. A horizontal tangent means the slope is 0.
Step 2: The curve is given by the equation y = x^3 - 3x + 2. We need to find the derivative of this function to determine the slope.
Step 3: Calculate the derivative f'(x) of the function y = x^3 - 3x + 2. The derivative is f'(x) = 3x^2 - 3.
Step 4: Set the derivative equal to 0 to find where the slope is horizontal: 3x^2 - 3 = 0.
Step 5: Solve the equation 3x^2 - 3 = 0. First, add 3 to both sides: 3x^2 = 3. Then, divide both sides by 3: x^2 = 1.
Step 6: Take the square root of both sides to find x: x = 1 or x = -1.
Step 7: Now, we need to find the corresponding y-coordinates for these x-values using the original equation y = x^3 - 3x + 2.
Step 8: For x = 1, substitute into the equation: y = 1^3 - 3(1) + 2 = 1 - 3 + 2 = 0. So, one point is (1, 0).
Step 9: For x = -1, substitute into the equation: y = (-1)^3 - 3(-1) + 2 = -1 + 3 + 2 = 4. So, another point is (-1, 4).
Step 10: The points on the curve where the tangent is horizontal are (1, 0) and (-1, 4).