Stationary Points and Function Shape
This topic uses differentiation as a tool to analyse function behaviour without plotting every point. By finding the derivative (gradient), you can pinpoint exactly where a graph has peaks and troughs (stationary points) and identify the intervals where it is rising or falling.
Part of the ESAT Mathematics 2 syllabus — revision for the Engineering and Science Admissions Test (ESAT), the UAT-UK admissions test for Cambridge, Imperial, Oxford and UCL.
Key points
- To find the x-coordinates of stationary points, calculate the first derivative (dy/dx), set it to zero, and solve the equation dy/dx = 0.
- A function is increasing when its gradient is positive (dy/dx > 0) and decreasing when its gradient is negative (dy/dx < 0).
- The nature of a stationary point can be determined using the second derivative test: if d2y/dx2 > 0, it is a local minimum; if d2y/dx2 < 0, it is a local maximum.
- Alternatively, you can test the sign of the first derivative (dy/dx) on either side of the stationary point. For a minimum, the gradient changes from negative to zero to positive.
- The regions where a function increases or decreases are the intervals between its stationary points. You can determine the behaviour in these intervals by testing a single value from each.
- Always find both the x and y coordinates if a question asks for the coordinates of a stationary point.
Diagram
Formulae
dy/dx = 0 To find the x-coordinates of stationary points.
dy/dx > 0 To find the interval(s) where the function is increasing.
dy/dx < 0 To find the interval(s) where the function is decreasing.
d2y/dx2 > 0 At a stationary point, this condition indicates a local minimum.
d2y/dx2 < 0 At a stationary point, this condition indicates a local maximum.
Definitions
- Stationary Point
- A point on a curve where the gradient is zero. At this point, the function is momentarily neither increasing nor decreasing. It can be a local maximum, local minimum, or a point of inflexion (though inflexions are excluded from this topic).
- Increasing Function
- A function whose value increases as the independent variable (x) increases. This corresponds to a positive gradient (dy/dx > 0).
- Decreasing Function
- A function whose value decreases as the independent variable (x) increases. This corresponds to a negative gradient (dy/dx < 0).
Worked example
Consider the function f(x) = 2x3 + 3x2 - 12x + 4. Find the coordinates of its local maximum point, and state the range of values of x for which the function is increasing.
- 1
First, find the derivative of the function:
f'(x) = 6x2 + 6x - 12 - 2
To find stationary points, set the derivative to zero:
6x2 + 6x - 12 = 0.
- 3
Simplify the equation by dividing by 6:
x2 + x - 2 = 0 - 4
Factorise the quadratic:
(x + 2)(x - 1) = 0.
The stationary points are at x = -2 and x = 1.
- 5
To determine their nature, find the second derivative:
f''(x) = 12x + 6 - 6 Test x = -2:f''(-2) = 12(-2) + 6 = -24 + 6 = -18
Since f''(-2) < 0, this is a local maximum.
- 7 Test x = 1:f''(1) = 12(1) + 6 = 18
Since f''(1) > 0, this is a local minimum.
- 8
The question asks for the coordinates of the local maximum, so substitute x = -2 into the original function:
f(-2) = 2(-2)3 + 3(-2)2 - 12(-2) + 4 = 2(-8) + 3(4) + 24 + 4 = -16 + 12 + 24 + 4 = 24The maximum is at (-2, 24).
- 9
The function is increasing when f'(x) > 0, i.e., 6x2 + 6x - 12 > 0.
This is a positive (U-shaped) parabola with roots at x = -2 and x = 1.
It is positive outside its roots.
- 10
Therefore, the function is increasing for x < -2 and x > 1.
Answer: The local maximum is at (-2, 24). The function is increasing for x < -2 or x > 1.
Common mistakes
- ×Simple arithmetic errors when solving the quadratic equation from dy/dx = 0, especially when factorising with negative numbers.
- ×Forgetting to substitute the x-coordinate back into the original function f(x), not the derivative f'(x), when finding the y-coordinate of a stationary point.
- ×Confusing the conditions for the second derivative test: thinking d2y/dx2 > 0 (positive) means a maximum, when it actually indicates a minimum (a 'positive' curvature, like a smile).
- ×When asked for an interval, incorrectly describing the region. For a positive quadratic derivative, the function increases *outside* the roots, not between them.
No-calculator tips
- ✓When solving dy/dx = 0 for a polynomial, always look for a common factor to divide out first. This makes the numbers smaller and factorisation easier.
- ✓If you forget the second derivative test, you can test the sign of the gradient (dy/dx) on either side of a stationary point. For a maximum at x=a, the gradient will be positive for x slightly less than a, and negative for x slightly more than a.
- ✓Sketch a quick graph of the derivative function, dy/dx. The intervals where this graph is above the x-axis are where the original function f(x) is increasing.