0580

Differentiation

Coordinate Geometry and Graphs

Where the gradient is zero

  • At the very top or the very bottom of a curve the tangent is horizontal, so the gradient there is zero
The same cubic with a short horizontal line drawn through each turning point, showing that the tangent there is flat and so the gradient is zero
Source: Finding stationary points & turning points by Save My Exams
  • Such a point is a turning point, also called a stationary point
A cubic curve with arrows along it showing the direction of travel and both turning points marked: at the first the curve is going up before and down after, and at the second it is going down before and up after
Source: Finding stationary points & turning points by Save My Exams
  • A maximum is a peak and a minimum is a trough, and both are found the same way
Two curves side by side, one with its maximum point marked at the top of a peak and one with its minimum point marked at the bottom of a trough
Source: Classifying stationary points by Save My Exams
  • The method is always the same three steps
    • Differentiate the equation of the curve
    • Put the derivative equal to zero, then solve that equation
    • Take each x found and substitute it into the curve, not the derivative, for the matching y
  • A cubic normally has two turning points, so expect two answers and give both as coordinates
  • Writing the line "dy/dx = 0" is itself worth a mark, so put it down before solving
Worked example

Finding both turning points of a cubic

Find the coordinates of the turning points of y = x³ − 27x + 4.

Solution:

  • A turning point is where the curve stops rising and starts falling, or the reverse, so the gradient there is zero
  • Differentiate first: dy/dx = 3x² − 27
  • Then set the derivative to zero — writing the line dy/dx = 0 earns a mark by itself
  • 3x² − 27 = 0
  • Divide through by 3: x² − 9 = 0, so x² = 9
  • Taking the square root gives two answers, x = 3 and x = −3, and missing the negative one loses half the question
  • Those are only the x coordinates; the y coordinates come from substituting back into the equation of the curve, never into the derivative
  • At x = 3: 3³ − 27 × 3 + 4 = 27 − 81 + 4 = −50
  • At x = −3: (−3)³ − 27 × (−3) + 4 = −27 + 81 + 4 = 58
  • The turning points are (3, −50) and (−3, 58)