Extrema Of A Function

Subsection 5.5.1 Relative Extrema

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A relative maximum point on a function is a point \((x,y)\) on the graph of the function whose \(y\)-coordinate is larger than all other \(y\)-coordinates on the graph at points “close to” \((x,y)\text{.}\) More precisely, \((x,f(x))\) is a relative maximum if there is an interval \((a,b)\) with \(a\lt x\lt b\) and \(f(x)\ge f(z)\) for every \(z\) in \((a,b)\text{.}\) Similarly, \((x,y)\) is a relative minimum point if it has locally the smallest \(y\)-coordinate. Again being more precise: \((x,f(x))\) is a relative minimum if there is an interval \((a,b)\) with \(a\lt x\lt b\) and \(f(x)\le f(z)\) for every \(z\) in \((a,b)\text{.}\) A relative extremum is either a relative minimum or a relative maximum.

Note:

1. The plural of extremum is extrema and similarly for maximum and minimum.

2. Because a relative extremum is “extreme” locally by looking at points “close to” it, it is also referred to as a local extremum.

Definition 5.46. Relative Maxima and Minima.

A real-valued function \(f\) has a relative maximum at \(x_0\) if \(f(x_0)\geq f(x)\) for all \(x\) in some open interval containing \(x_0\text{.}\)

A real-valued function \(f\) has a relative minimum at \(x_0\) if \(f(x_0)\leq f(x)\) for all \(x\) in some open interval containing \(x_0\text{.}\)

Relative maximum and minimum points are quite distinctive on the graph of a function, and are therefore useful in understanding the shape of the graph. In many applied problems we want to find the largest or smallest value that a function achieves (for example, we might want to find the minimum cost at which some task can be performed) and so identifying maximum and minimum points will be useful for applied problems as well. Some examples of relative maximum and minimum points are shown in Figure 5.14.

If \((x,f(x))\) is a point where \(f(x)\) reaches a relative maximum or minimum, and if the derivative of \(f\) exists at \(x\text{,}\) then the graph has a tangent line and the tangent line must be horizontal. This is important enough to state as a theorem.

The proof is simple enough and we include it here, but you may accept Fermat's Theorem based on its strong intuitive appeal and come back to its proof at a later time.

Theorem 5.47. Fermat's Theorem.

If \(f(x)\) has a relative extremum at \(x=a\) and \(f\) is differentiable at \(a\text{,}\) then \(f'(a)=0\text{,}\) provided that \(f'(a)\) exists.

Proof.

We shall give the proof for the case where \(f\left( x\right)\) has a relative maximum at \(x=a\text{.}\) The proof for the relative minimum case is similar.

Since \(f\left( x\right)\) has a relative maximum at \(x=a\text{,}\) there is an open interval \(\left( c,d\right)\) with \(c\lt a\lt d\) and \(f\left( x\right) \leq f\left( a\right)\) for every \(x\) in \(\left( c,d\right) \text{.}\) So, \(f\left( x\right) -f\left( a\right) \leq 0\) for all such \(x\text{.}\) Let us now look at the sign of the difference quotient \(\dfrac{f\left( x\right) -f\left( a\right) }{x-a}\text{.}\) We consider two cases according as \(x>a\) or \(x\lt a\text{.}\)

If \(x>a\text{,}\) then \(x-a>0\) and so, \(\dfrac{f\left( x\right) -f\left( a\right) }{x-a}\leq 0\text{.}\) Taking limit as \(x\) approach \(a\) from the right, we get

\begin{equation*} \lim_{x\rightarrow a^{+}}\frac{f\left( x\right) -f\left( a\right) }{x-a}\leq 0\text{.} \end{equation*}

On the other hand, if \(x\lt a\text{,}\) then \(x-a\lt 0\) and so,\(\dfrac{f\left( x\right) -f\left( a\right) }{x-a}\geq 0\text{.}\) Taking limit as \(x\) approach \(a\) from the left, we get

\begin{equation*} \lim_{x\rightarrow a^{-}}\frac{f\left( x\right) -f\left( a\right) }{x-a}\geq 0\text{.} \end{equation*}

Since \(f\) is differentiable at \(a\text{,}\)

\begin{equation*} f^{\prime }\left( a\right) =\lim\limits_{x\rightarrow a^{+}}\dfrac{f\left( x\right) -f\left( a\right) }{x-a}=\lim\limits_{x\rightarrow a^{-}}\dfrac{f\left( x\right) -f\left( a\right) }{x-a}\text{.} \end{equation*}

Therefore, we have both \(f^{\prime }\left( a\right) \leq 0\) and \(f^{\prime }\left( a\right) \geq 0\text{.}\) So, \(f^{\prime }\left( a\right) =0\text{.}\)

Thus, the only points at which a function can have a relative maximum or minimum are points at which the derivative is zero, as in the left-hand graph in Figure 5.14, or the derivative is undefined, as in the right-hand graph. This leads us to define these special points.

Definition 5.48. Critical Point.

Any value of \(x\) in the domain of \(f\) for which \(f'(x)\) is zero or undefined is called a critical point of \(f\text{.}\)

The \(x\)-values \(a\text{,}\) \(b\) and \(c\) above are places for which \(f'(x)\) is zero, and the \(x\)-values \(d\) and \(e\) above are places for which \(f'(x)\) is undefined.

Note: When looking for relative maximum and minimum points, you are likely to make two sorts of mistakes.

1. You may forget that a maximum or minimum can occur where the derivative does not exist. You should therefore check whether the derivative exists everywhere.

2. You might also assume that any place that the derivative is zero is a relative maximum or minimum point, but this is not true. A portion of the graph of \(\ds f(x)=x^3\) is shown in Figure 5.15. The derivative of \(f\) is \(f'(x)=3x^2\text{,}\) and \(f'(0)=0\text{,}\) but there is neither a maximum nor minimum at \((0,0)\text{.}\) In other words, the converse of Fermat's Theorem — if \(f'(a) = 0\) at some point \(x=a\text{,}\) then \(f\) must have a relative extremum at that point — is not true.

Since the derivative is zero or undefined at both relative maximum and relative minimum points, we need a way to determine which, if either, actually occurs. The most elementary approach, but one that is often tedious or difficult, is to test directly whether the \(y\)-coordinates “near” the potential maximum or minimum are above or below the \(y\)-coordinate at the point of interest. Of course, there are too many points “near” the point to test, but a little thought shows we need only test two provided we know that \(f\) is continuous (recall that this means that the graph of \(f\) has no jumps or gaps).

Suppose, for example, that we have identified three points at which \(f'\) is zero or nonexistent: \(\ds (x_1,y_1)\text{,}\) \(\ds (x_2,y_2)\text{,}\) \(\ds (x_3,y_3)\text{,}\) and \(\ds x_1\lt x_2\lt x_3\) (see Figure 5.16). Suppose that we compute the value of \(f(a)\) for \(\ds x_1\lt a\lt x_2\text{,}\) and that \(\ds f(a)\lt f(x_2)\text{.}\) What can we say about the graph between \(a\) and \(\ds x_2\text{?}\) Could there be a point \(\ds (b,f(b))\text{,}\) \(\ds a\lt b\lt x_2\) with \(\ds f(b)>f(x_2)\text{?}\) No: if there were, the graph would go up from \((a,f(a))\) to \((b,f(b))\) then down to \(\ds (x_2,f(x_2))\) and somewhere in between would have a relative maximum point. (This is not obvious; it is a result of the Extreme Value Theorem stated in the next section.) But at that relative maximum point the derivative of \(f\) would be zero or nonexistent, yet we already know that the derivative is zero or nonexistent only at \(\ds x_1\text{,}\) \(\ds x_2\text{,}\) and \(\ds x_3\text{.}\) The upshot is that one computation tells us that \(\ds (x_2,f(x_2))\) has the largest \(y\)-coordinate of any point on the graph near \(\ds x_2\) and to the left of \(\ds x_2\text{.}\) We can perform the same test on the right. If we find that on both sides of \(\ds x_2\) the values are smaller, then there must be a relative maximum at \(\ds (x_2,f(x_2))\text{;}\) if we find that on both sides of \(\ds x_2\) the values are larger, then there must be a relative minimum at \(\ds (x_2,f(x_2))\text{;}\) if we find one of each, then there is neither a relative maximum or minimum at \(\ds x_2\text{.}\)

It is not always easy to compute the value of a function at a particular point. The task is made easier by the availability of calculators and computers, but they have their own drawbacks—they do not always allow us to distinguish between values that are very close together. Nevertheless, because this method is conceptually simple and sometimes easy to perform, you should always consider it.

Example 5.49. Testing for Relative Extrema in Cubic Function.

Find all relative maximum and minimum points for the function \(\ds f(x)=x^3-x\text{.}\)

Solution

The derivative is \(\ds f'(x)=3x^2-1\text{.}\) This is defined everywhere and is zero at \(\ds x=\pm \sqrt{3}/3\text{.}\) Looking first at \(\ds x=\sqrt{3}/3\text{,}\) we see that \(\ds f(\sqrt{3}/3)=-2\sqrt{3}/9\text{.}\) Now we test two points on either side of \(\ds x=\sqrt{3}/3\text{,}\) choosing one point in the interval \((-\sqrt{3}/3,\sqrt{3}/3)\) and one point in the interval \((\sqrt{3}/3,\infty)\text{.}\) Since \(\ds f(0)=0>-2\sqrt{3}/9\) and \(\ds f(1)=0>-2\sqrt{3}/9\text{,}\) there must be a relative minimum at \(\ds x=\sqrt{3}/3\text{.}\) For \(\ds x=-\sqrt{3}/3\text{,}\) we see that \(\ds f(-\sqrt{3}/3)=2\sqrt{3}/9\text{.}\) This time we can use \(x=0\) and \(x=-1\text{,}\) and we find that \(\ds f(-1)=f(0)=0\lt 2\sqrt{3}/9\text{,}\) so there must be a relative maximum at \(\ds x=-\sqrt{3}/3\text{.}\)

Of course this example is made very simple by our choice of points to test, namely \(x=-1\text{,}\) \(0\text{,}\) \(1\text{.}\) We could have used other values, say \(-5/4\text{,}\) \(1/3\text{,}\) and \(3/4\text{,}\) but this would have made the calculations considerably more tedious, and we should always choose very simple points to test if we can.

Example 5.50. Testing for Relative Extrema in Trigonometric Function.

Find all relative maximum and minimum points for \(f(x)=\sin x+\cos x\text{.}\)

Solution

The derivative is \(f'(x)=\cos x-\sin x\text{.}\) This is always defined and is zero whenever \(\cos x=\sin x\text{.}\) Recalling that the \(\cos x\) and \(\sin x\) are the \(x\)- and \(y\)-coordinates of points on a unit circle, we see that \(\cos x=\sin x\) when \(x\) is \(\pi/4\text{,}\) \(\pi/4\pm\pi\text{,}\) \(\pi/4\pm2\pi\text{,}\) \(\pi/4\pm3\pi\text{,}\) etc. Since both sine and cosine have a period of \(2\pi\text{,}\) we need only determine the status of \(x=\pi/4\) and \(x=5\pi/4\text{.}\) We can use \(0\) and \(\pi/2\) to test the critical value \(x= \pi/4\text{.}\) We find that \(\ds f(\pi/4)=\sqrt{2}\text{,}\) \(\ds f(0)=1\lt \sqrt{2}\) and \(\ds f(\pi/2)=1\text{,}\) so there is a relative maximum when \(x=\pi/4\) and also when \(x=\pi/4\pm2\pi\text{,}\) \(\pi/4\pm4\pi\text{,}\) etc. We can summarize this more neatly by saying that there are relative maxima at \(\pi/4\pm 2k\pi\) for every integer \(k\text{.}\)

We use \(\pi\) and \(\frac{3\pi}{2}\) to test the critical value \(x=5\pi/4\text{.}\) The relevant values are \(\ds f(5\pi/4)=-\sqrt2\text{,}\) \(\ds f(\pi)=-1>-\sqrt2\text{,}\) \(\ds f(3\pi/2)=1>-\sqrt2\text{,}\) so there is a relative minimum at \(x=5\pi/4\text{,}\) \(5\pi/4\pm2\pi\text{,}\) \(5\pi/4\pm4\pi\text{,}\) etc. More succinctly, there are relative minima at \(5\pi/4\pm 2k\pi\) for every integer \(k\text{.}\)

Example 5.51. Testing for Relative Extrema in Power Function.

Find all relative maximum and minimum points for \(g\left( x\right) =x^{2/3}\text{.}\)

Solution

The derivative is \(g^{\prime }\left( x\right) =\frac{2}{3}x^{-1/3}\text{.}\) This is undefined when \(x=0\) and is not equal to zero for any \(x\) in the domain of \(g^{\prime }\text{.}\) Now we test two points on either side of \(x=0\text{.}\) We use \(x=-1\) and \(x=1\text{.}\) Since \(g\left( 0\right) =0\text{,}\) \(g\left( -1\right) =1>0\) and \(g\left( 1\right) =1>0\text{,}\) there must be a relative minimum at \(x=0\text{.}\)