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

Practice Questions

Q1
Find the coordinates of the point on the curve y = x^3 - 3x + 2 where the slope of the tangent is 0.
  1. (1, 0)
  2. (0, 2)
  3. (2, 0)
  4. (3, 2)

Questions & Step-by-Step Solutions

Find the coordinates of the point on the curve y = x^3 - 3x + 2 where the slope of the tangent is 0.
Correct Answer: (1, 0)
  • 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).
  • Derivative and Slope of Tangent – Understanding how to find the slope of a curve at a given point using derivatives.
  • Finding Critical Points – Identifying points where the derivative equals zero to find local maxima, minima, or points of inflection.
  • Evaluating Functions – Calculating the value of the original function at specific x-coordinates to find corresponding y-coordinates.
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