Find the coordinates of the point on the curve y = x^3 - 3x + 2 where the slope of the tangent is 0.

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Q: Find the coordinates of the point on the curve y = x^3 - 3x + 2 where the slope of the tangent is 0.

Solution: f'(x) = 3x^2 - 3. Setting f'(x) = 0 gives x^2 = 1, so x = 1 or x = -1. f(1) = 0, f(-1) = 4. The point is (1, 0).

Steps: 10

Step 1: Start with the given curve equation, which is y = x^3 - 3x + 2.
Step 2: To find where the slope of the tangent is 0, we need to find the derivative of the function. The derivative f'(x) represents the slope.
Step 3: Calculate the derivative of the function: f'(x) = 3x^2 - 3.
Step 4: Set the derivative equal to 0 to find the points where the slope is 0: 3x^2 - 3 = 0.
Step 5: Simplify the equation: 3x^2 = 3, then divide both sides by 3: x^2 = 1.
Step 6: Solve for x by taking the square root of both sides: x = 1 or x = -1.
Step 7: Now, we need to find the corresponding y-coordinates for both x-values by substituting them back into the original equation.
Step 8: For x = 1, calculate y: f(1) = 1^3 - 3(1) + 2 = 1 - 3 + 2 = 0.
Step 9: For x = -1, calculate y: f(-1) = (-1)^3 - 3(-1) + 2 = -1 + 3 + 2 = 4.
Step 10: We have two points: (1, 0) and (-1, 4). The point where the slope of the tangent is 0 is (1, 0).