Stationary points

The stationary points of a graph \(y=f(x)\) are those points \((x,y)\) on the graph where \(f'(x) = 0\). A stationary point can be a turning point or a stationary point of inflexion.

Differentiating the term \(a_k x^k\) in a polynomial gives \(k a_k x^{k-1}\). So if a polynomial \(f(x)\) has degree \(n\), then its derivative \(f'(x)\) has degree \(n-1\). To find stationary points of \(y=f(x)\), we must solve the polynomial equation \(f'(x)=0\) of degree \(n-1\).

Take an example from our gallery.


Let \(f(x) = 2x^3 - 3x^2 + 5\). Find the stationary points of the graph \(y=f(x)\).


We compute \(f'(x) = 6x^2 - 6x\). To find stationary points we solve \(6x^2 - 6x\) = 0. Factorising to \(6x(x-1)=0\) gives \(x=0\) or \(x = 1\). Substituting these values of \(x\) gives \(f(0) = 5\) and \(f(1) = 4\). So the stationary points are \((0,5)\) and \((1,4)\).

Note. Here \(f\) has degree 3, its derivative \(f'\) has degree 2, and so \(f'(x)=0\) is a quadratic equation.

The following exercise shows that a polynomial graph may have no stationary points. In this exercise, the polynomial \(f\) again has degree 3 and its derivative \(f'\) has degree 2, but the equation \(f'(x) = 0\) has no solutions.

Exercise 8

Let \(f(x) = x^3 + x - 2\). Show that \(f(x)\) has no stationary points.

Screencast of exercise 8 mp4 of screencast of exercise 8

Next let's consider the number of stationary points of a polynomial graph \(y=f(x)\), where \(f(x)\) has degree \(n\).

The stationary points are found by solving the equation \(f'(x)=0\), which has degree \({n-1}\), and hence has at most \(n-1\) real solutions. Therefore the graph \(y=f(x)\) has at most \(n-1\) stationary points. This confirms our Conjecture 3.

Slightly more trickily, if the degree \(n\) is even, then the degree \(n - 1\) of the derivative \(f'(x)\) is odd. So the graph of \(f'(x)\) goes from \(-\infty\) to \(+\infty\), or vice versa. Therefore the graph of \(f'(x)\) crosses the \(x\)-axis somewhere, changing sign from positive to negative or from negative to positive. This gives a turning point of \(f(x)\). We have now confirmed Conjecture 4: when \(n\) is even, the graph of \(f(x)\) has at least one turning point.

The next exercise gives a test for the number of stationary points of a cubic polynomial.

Exercise 9
Let \(f(x) = ax^3 + bx^2 + cx + d\), where \(a,b,c,d\) are real numbers with \(a \ne 0\). Show that:
  1. If \(b^2 - 3ac < 0\), then \(y=f(x)\) has no stationary points.
  2. If \(b^2 - 3ac = 0\), then \(y=f(x)\) has one stationary point.
  3. If \(b^2 - 3ac > 0\), then \(y=f(x)\) has two distinct stationary points.

Screencast of interactive 3 mp4 of interactive CDF 3,  Interactive 3 CDF of interactive 3

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