Critical Points


We will discuss the occurrence of local maxima and local minima of a function. In fact, these points are crucial to many questions related to optimization problems. We will discuss these problems in later pages.

Definition. A function f(x) is said to have a local maximum at c iff there exists an interval I around c such that

\begin{displaymath}f(c) \geq f(x) \;\;\mbox{for all $x \in I$}.\end{displaymath}

Analogously, f(x) is said to have a local minimum at c iff there exists an interval I around c such that

\begin{displaymath}f(c) \leq f(x) \;\;\mbox{for all $x \in I$}.\end{displaymath}

A local extremum is a local maximum or a local minimum.

Using the definition of the derivative, we can easily show that:

If f(x) has a local extremum at c, then either

\begin{displaymath}f'(c) = 0\;\;\mbox{or $f'(c)$ does not exist.}\end{displaymath}

These points are called critical points.

Example. Consider the function f(x) = x3. Then f'(0) = 0 but 0 is not a local extremum. Indeed, if x < 0, then f(x) < f(0) and if x > 0, then f(x) > f(0).

Therefore the conditions

\begin{displaymath}f'(c) = 0\;\;\mbox{or $f'(x)$ does not exist.}\end{displaymath}

do not imply in general that c is a local extremum. So a local extremum must occur at a critical point, but the converse may not be true.

Example. Let us find the critical points of

f(x) = |x2-x|

Answer. We have

\begin{displaymath}f(x) = \left\{\begin{array}{clcll}
x^2-x &\mbox{if}& \;\; x \...
...leq 1\\
x^2-x &\mbox{if}& \;\; 1 \leq x.\\
\end{array}\right.\end{displaymath}

Clearly we have

\begin{displaymath}f'(x) = \left\{\begin{array}{clcll}
2x-1 &\mbox{if}& \;\; x <...
... 0< x < 1\\
2x-1 &\mbox{if}& \;\; 1 < x.\\
\end{array}\right.\end{displaymath}

Clearly we have

\begin{displaymath}f'(x) = 0 \;\;\mbox{iff}\;\; x = \frac{1}{2}\cdot\end{displaymath}

Also one may easily show that f'(0) and f'(1) do not exist. Therefore the critical points are

\begin{displaymath}\frac{1}{2},\;0,\;1.\end{displaymath}


Let c be a critical point for f(x). Assume that there exists an interval I around c, that is c is an interior point of I, such that f(x) is increasing to the left of c and decreasing to the right, then c is a local maximum. This implies that if $f'(x) \geq 0$ for $x \leq c$ (x close to c), and $f'(x) \leq 0$ for $x \geq c$ (x close to c), then c is a local maximum. Note that similarly if $f'(x) \leq 0$ for $x \leq c$ (x close to c), and $f'(x) \geq 0$ for $x \geq c$(x close to c), then c is a local minimum.

So we have the following result:

First Derivative Test. If c is a critical point for f(x), such that f '(x) changes its sign as x crosses from the left to the right of c, then c is a local extremum.

Example. Find the local extrema of

f(x) = |x2-x|

Answer. Since the local extrema are critical points, then from the above discussion, the local extrema, if they exist, are among the points

\begin{displaymath}\frac{1}{2},\;0,\;1.\end{displaymath}

Recall that

\begin{displaymath}f'(x) = \left\{\begin{array}{clcll}
2x-1 &\mbox{if}& \;\; x <...
... 0< x < 1\\
2x-1 &\mbox{if}& \;\; 1 < x.\\
\end{array}\right.\end{displaymath}

(1)
For x = 1/2, we have

\begin{displaymath}\left\{\begin{array}{clcll}
f'(x) > 0 &\mbox{if}& \;\; 0 < x ...
...
f'(x) < 0 &\mbox{if}& \;\; 1/2 < x < 1.\\
\end{array}\right.\end{displaymath}

So the critical point $\frac{1}{2}$ is a local maximum.
(2)
For x = 0, we have

\begin{displaymath}\left\{\begin{array}{clcll}
f'(x) < 0 &\mbox{if}& \;\; x <0\\
f'(x) > 0 &\mbox{if}& \;\; 0< x < 1/2.\\
\end{array}\right.\end{displaymath}

So the critical point 0 is a local minimum.
(3)
For x = 1, we have

\begin{displaymath}\left\{\begin{array}{clcll}
f'(x) < 0 &\mbox{if}& \;\; 1/2 < x <1\\
f'(x) > 0 &\mbox{if}& \;\; 1 < x .\\
\end{array}\right.\end{displaymath}

So the critical point -1 is a local minimum.


Let c be a critical point for f(x) such that f'(c) =0.
(i)
If f''(c) > 0, then f'(x) is increasing in an interval around c. Since f'(c) =0, then f'(x) must be negative to the left of c and positive to the right of c. Therefore, c is a local minimum.
(ii)
If f''(c) < 0, then f'(x) is decreasing in an interval around c. Since f'(c) =0, then f'(x) must be positive to the left of c and negative to the right of c. Therefore, c is a local maximum.
This test is known as the Second-Derivative Test.

Example. Find the local extrema of

f(x) = x5 - 5 x.

Answer. First let us find the critical points. Since f(x) is a polynomial function, then f(x) is continuous and differentiable everywhere. So the critical points are the roots of the equation f'(x) = 0, that is 5x4 - 5 = 0, or equivalently x4 - 1 =0. Since x4 - 1 = (x-1)(x+1)(x2+1), then the critical points are 1 and -1. Since f''(x) = 20 x3, then

\begin{displaymath}f''(1) = 20 > 0 \;\;\mbox{and}\;\; f''(-1) = -20 < 0.\end{displaymath}

The second-derivative test implies that x=1 is a local minimum and x= -1 is a local maximum.


Exercise 1. Find the local extrema of

\begin{displaymath}f(x) = \frac{x}{1+x^2}\cdot\end{displaymath}

Answer.

Exercise 2. Find the local extrema of

\begin{displaymath}f(x) = \sin(x) + \cos(x).\end{displaymath}

Answer.


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