Application to finding extremal points.

The following proposition is often taught without proof. A nice proof is writtenusing the Taylor developments of the function $ f$ at $ x_0$ (cf Chapter 10 section Taylor series).

Proposition 6.10.11   Let $ f$ be a function defined on a neighborhood of $ x_0$ . Suppose that $ f$ is differentiable (at least) twice at $ x_0$ and that $ f'(x_0) =0$ .
  1. If $ f''(x_0)<0$ , then $ f$ has a maximum at $ x_0$ .
  2. If $ f''(x_0)>0$ , then $ f$ has a minimum at $ x_0$ .

If $ f''(x_0)=0$ , then the behavior of the function $ f$ at $ x_0$ depends on the derivatives of higher order.

Proposition 6.10.12   Let $ f$ be a function defined on a neighborhood of $ x_0$ . Suppose that $ f$ has enough derivatives on this neighborhood and that $ f'(x_0)=f''(x_0)=0$ . Denote by $ n$ the order of the first derivative which does not vanish at $ x_0$ .
i.
If $ n$ even and $ f^{(n)}(x_0)<0$ , then $ f$ has an maximum at $ x_0$ .
ii.
If $ n$ even and $ f^{(n)}(x_0)>0$ , then $ f$ has an minimum at $ x_0$ .
iii.
If $ n$ odd, then $ f$ has a point of inflexion at $ x_0$ .

Example 6.10.13       

  1. Let $ f(x)=x^4$ . Then $ f'(x)=4x^3$ , $ f''(x)=12x^2$ . Both derivatives vanish at 0. The first occurrence of a derivative which does not vanish at 0 is the $ 4^{th}$ derivative and $ f^{(4)}(0)=24$ . It follows that $ f$ has a minimum at 0 (see Figure 13(a)).
  2. Let $ f(x)=x^5$ . Then $ f'(x)=5x^4$ , $ f''(x)=20x^3$ . Both derivatives vanish at 0. The first occurrence of a derivative which does not vanish at 0 is the $ 5^{th}$ derivative. This order is odd, hence $ f$ has a point of inflexion at 0 (see Figure 13(b)).

Figure 13:
\begin{figure}\centering
\mbox{
\subfigure[$f(x)=x^4$.]{\epsfig{file=x4.eps,heig...
...\qquad
\subfigure[$f(x)=x^5$.]{\epsfig{file=x5.eps,height=5cm}}
}\end{figure}

Example 6.10.14   Let $ f(x)=\sin x + \sin 2x$ ; the graph of $ f$ is displayed on Figure 14.

The function $ f$ is defined by a trigonometric polynomial, hence it is differentiable over $ \mathbb{R}$ . We have: $ f'(x)= \cos x + 2 \cos 2x$ . Now check that this first derivative vanishes at $ x_0 = -0.935929 \;$   rd .

Figure 14: Extrema of a trigonometric function.
\begin{figure}\centering
\mbox{\epsfig{file=SinxPlusSin2x.eps,height=6cm}}\end{figure}

We have: $ f''(x)= -\sin x - 4 \sin 2x$ ; then $ f''(x_0)>0$ . Therefore the function $ f$ has a minimum at $ x_0$ .Similarly, we see that $ f'$ vanishes at $ x_1=0.935929 \;$   rd ; there we have $ f''(x_0)<0$ , i.e. $ f$ has a maximum at $ x_1$ .

Of course, the above computations are not sufficient for deciding whether these extrema are absolute or local.

Noah Dana-Picard 2007-12-28