4.3 Maxima and minima  (Page 3/15)

To answer this question, let’s look at [link] again. The local extrema occur at $x=0,$ $x=1,$ and $x=2.$ Notice that at $x=0$ and $x=1,$ the derivative $f\prime (x)=0.$ At $x=2,$ the derivative $f\prime \left(x\right)$ does not exist, since the function $f$ has a corner there. In fact, if $f$ has a local extremum at a point $x=c,$ the derivative $f\prime (c)$ must satisfy one of the following conditions: either $f\prime (c)=0$ or $f\prime \left(c\right)$ is undefined. Such a value $c$ is known as a critical point and it is important in finding extreme values for functions.

As mentioned earlier, if $f$ has a local extremum at a point $x=c,$ then $c$ must be a critical point of $f.$ This fact is known as Fermat’s theorem    .

Proof

Suppose $f$ has a local extremum at $c$ and $f$ is differentiable at $c.$ We need to show that $f\prime \left(c\right)=0.$ To do this, we will show that $f\prime \left(c\right)\ge 0$ and $f\prime \left(c\right)\le 0,$ and therefore $f\prime \left(c\right)=0.$ Since $f$ has a local extremum at $c,$ $f$ has a local maximum or local minimum at $c.$ Suppose $f$ has a local maximum at $c.$ The case in which $f$ has a local minimum at $c$ can be handled similarly. There then exists an open interval $I$ such that $f\left(c\right)\ge f\left(x\right)$ for all $x\in I.$ Since $f$ is differentiable at $c,$ from the definition of the derivative, we know that

Since this limit exists, both one-sided limits also exist and equal $f\prime (c).$ Therefore,

and

Since $f(c)$ is a local maximum, we see that $f(x)-f(c)\le 0$ for $x$ near $c.$ Therefore, for $x$ near $c,$ but $x>c,$ we have $\frac{f\left(x\right)-f\left(c\right)}{x-c}\le 0.$ From [link] we conclude that $f\prime (c)\le 0.$ Similarly, it can be shown that $f\prime \left(c\right)\ge 0.$ Therefore, $f\prime (c)=0.$

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From Fermat’s theorem, we conclude that if $f$ has a local extremum at $c,$ then either $f\prime \left(c\right)=0$ or $f\prime \left(c\right)$ is undefined. In other words, local extrema can only occur at critical points.

Note this theorem does not claim that a function $f$ must have a local extremum at a critical point. Rather, it states that critical points are candidates for local extrema. For example, consider the function $f\left(x\right)=x^3.$ We have $f\prime \left(x\right)=3x^2=0$ when $x=0.$ Therefore, $x=0$ is a critical point. However, $f(x)=x^3$ is increasing over $(\text{−}\infty ,\infty ),$ and thus $f$ does not have a local extremum at $x=0.$ In [link] , we see several different possibilities for critical points. In some of these cases, the functions have local extrema at critical points, whereas in other cases the functions do not. Note that these graphs do not show all possibilities for the behavior of a function at a critical point.

Later in this chapter we look at analytical methods for determining whether a function actually has a local extremum at a critical point. For now, let’s turn our attention to finding critical points. We will use graphical observations to determine whether a critical point is associated with a local extremum.