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Stationary Points, Increasing and Decreasing Functions

How to find and classify stationary points, identify increasing and decreasing intervals, and use this information to sketch curves.

What a stationary point is

At a stationary point, the gradient is zero and the tangent is horizontal. Find candidates by solving \frac{dy}{dx}=0. Then substitute each x-value into the original equation to find its y-coordinate.

A local maximum is higher than nearby points; a local minimum is lower than nearby points. These need not be the greatest or least values over the whole domain.

Classifying stationary points

At a stationary point, a positive second derivative means a minimum, and a negative second derivative means a maximum. If \frac{d^2y}{dx^2}=0, this test does not decide the nature.

You can instead check the sign of the first derivative on either side:

  • Positive to negative: maximum.
  • Negative to positive: minimum.
  • No sign change: neither a maximum nor a minimum.

Worked example 1

Find the stationary point of y=2x^2-12x+11 and determine its nature.

  1. Differentiate: \frac{dy}{dx}=4x-12. Set this equal to zero.

  2. Your turn. Solve 4x-12=0

    x=3

  3. Use the original curve: y=2(3)^2-12(3)+11=-7
  4. \frac{d^2y}{dx^2}=4>0, so the point is a minimum.
  5. Answer: a minimum at (3,-7)

Worked example 2

Find and classify both stationary points of y=x^3-3x^2-9x+5.

  1. \frac{dy}{dx}=3x^2-6x-9=3(x-3)(x+1)

  2. Your turn. Solve 3(x-3)(x+1)=0. Enter both x-values.

    Give every value, separated by commas

    x=-1 or x=3

  3. At x=-1, y=10. At x=3, y=-22
  4. \frac{d^2y}{dx^2}=6x-6. Its values are -12 at x=-1 and 12 at x=3
  5. Answer: a local maximum at (-1,10) and a local minimum at (3,-22)

Increasing and decreasing functions

Where \frac{dy}{dx}>0, the curve rises from left to right. Where \frac{dy}{dx}<0, it falls. To find these intervals, solve an inequality involving the derivative, not the original function.

The sign of y tells you whether a point is above or below the x-axis. It does not tell you whether the function is increasing.

Explore the signs of the gradient

Try it: move through x=-1 and x=1. The gradient signs go positive, negative, positive. The two stationary points separate the increasing and decreasing intervals.

Worked example 3

For y=x^3-3x^2-9x+5, find the intervals where the derivative is positive and where it is negative.

  1. From worked example 2, \frac{dy}{dx}=3(x-3)(x+1), with zeros at -1 and 3
  2. Test one value in each interval: at x=-2 the derivative is 15; at x=0 it is -9; at x=4 it is 15

  3. Your turn. Write the single double inequality for the interval where the derivative is negative.

    Type powers with ^ and fractions with /, for example x^2 or (x+1)/3

    The derivative is negative for -1<x<3

  4. Answer: positive for x<-1 or x>3, and negative for -1<x<3. The derivative equals zero at the two boundary values.

When the second derivative is zero

A zero second derivative is inconclusive. Check the first derivative just before and after the stationary point. Do not automatically call it a maximum, a minimum or neither.

For comparison, y=x^3+2 has derivative 3x^2. Its stationary point (0,2) has positive gradient on both sides, so it is neither a maximum nor a minimum. An isolated zero gradient need not stop a function being increasing across an interval.

Worked example 4

Find the stationary point of y=x^4+3 and use gradient signs to determine its nature.

  1. \frac{dy}{dx}=4x^3=0 gives x=0, so the stationary point is (0,3)
  2. \frac{d^2y}{dx^2}=12x^2 is also zero there, so use the first derivative on either side.

  3. Your turn. Work out 4x^3 when x=-1

    At x=-1, the gradient is -4

  4. For every x<0, 4x^3<0; for every x>0, 4x^3>0. The sign changes from negative to positive.
  5. Answer: a minimum at (0,3)

Worked example 5

Show that y=x^3+3x^2+5x-2 is increasing for all real x.

  1. Differentiate: \frac{dy}{dx}=3x^2+6x+5

  2. Your turn. Write 3x^2+6x+5 in completed square form.

    Type powers with ^ and fractions with /, for example x^2 or (x+1)/3

    3x^2+6x+5=3(x+1)^2+2

  3. Since (x+1)^2\ge0, the derivative is at least 2 and is always positive.
  4. Answer: the derivative is positive for every real x, so the function is increasing throughout its domain.

Using stationary points in a sketch

Mark stationary points, any easily found intercepts, and the general behaviour at the ends. Join them with a smooth curve, following the increasing and decreasing intervals.

Worked example 6

Find the stationary points and intercepts needed to sketch y=x^3-3x^2. Describe the shape of the sketch.

  1. \frac{dy}{dx}=3x(x-2), so the stationary x-values are 0 and 2

  2. Your turn. Find y when x=2 in the original curve.

    y=2^3-3(2)^2=-4

  3. \frac{d^2y}{dx^2}=6x-6: (0,0) is a local maximum and (2,-4) is a local minimum.
  4. For the x-intercepts, x^2(x-3)=0, giving (0,0) and (3,0). The origin is also the y-intercept.
  5. Answer: the curve rises from the bottom left to the maximum (0,0), touches the axis there, falls to (2,-4), then rises through (3,0) towards the top right.

Common mistakes

  • Solving for intercepts instead of stationary points. Solve \frac{dy}{dx}=0, not y=0.
  • Using the derivative for a point’s height. Substitute its input into the original function.
  • Reversing the second derivative test. At a stationary point, positive means minimum; negative means maximum.
  • Treating a zero second derivative as conclusive. Check the first derivative’s sign on either side.
  • Using height to decide whether a curve increases. Use the sign of the first derivative, not the sign of the function.
  • Calling a local maximum the greatest value. Check the whole domain, including any included endpoints.

Now try it: Stationary Points, Increasing and Decreasing Functions practice questions

More on this topic: Stationary Points, Increasing and Decreasing Functions worksheet with full solutions