1.9: Increasing, Decreasing, and Local Extrema

Recall that the slope of a line is positive if, and only if, the line rises from left to right. That is, if \(m>0, f(x)=m x+b,\) and \(u<v,\) then

\[\begin{aligned} f(v) &=m v+b \\ &=m v-m u+m u+b \\ &=m(v-u)+m u+b \\ &>m u+b \\ &=f(u) . \end{aligned}\] We should expect that an analogous statement holds for differentiable functions: if \(f\) is differentiable and \(f^{\prime}(x)>0\) for all \(x\) in an interval \((a, b),\) then \(f(v)>f(u)\) for any \(v>u\) in \((a, b) .\) This is in fact the case, although the inference requires establishing a direct connection between slope at a point and the average slope over an interval, or, in terms of rates of change, between the instantaneous rate of change at a point and the average rate of change over an interval. The mean-value theorem makes this connection.

The Mean-Value Theorem

Recall that the extreme value property tells us that a continuous function on a closed interval must attain both a minimum and a maximum value. Suppose \(f\) is continuous on \([a, b],\) differentiable on \((a, b),\) and \(f\) attains a maximum value at \(c\) with \(a<c<b .\) In particular, for any infinitesimal \(d x, f(c) \geq f(c+d x),\) and so, equivalently, \(f(c+d x)-f(c) \leq 0 .\) It follows that if \(d x>0,\)

\[\frac{f(c+d x)-f(c)}{d x} \leq 0 ,\] and if \(d x<0\), \[\frac{f(c+d x)-f(c)}{d x} \geq 0 .\] Since both of these values must be infinitesimally close to the same real number, it must be the case that \[\frac{f(c+d x)-f(c)}{d x} \simeq 0 .\] That is, we must have \(f^{\prime}(c)=0 .\) A similar result holds if \(f\) has a minimum at \(c,\) and so we have the following basic result.

Theorem \(\PageIndex{1}\)

If \(f\) is differentiable on \((a, b)\) and attains a maximum, or a minimum, value at \(c,\) then \(f^{\prime}(c)=0\).

Now suppose \(f\) is continuous on \([a, b],\) differentiable on \((a, b),\) and \(f(a)=f(b) .\) If \(f\) is a constant function, then \(f^{\prime}(c)=0\) for all \(c\) in \((a, b) .\) If \(f\) is not constant, then there is a point \(c\) in \((a, b)\) at which \(f\) attains either a maximum or a minimum value, and so \(f^{\prime}(c)=0 .\) In either case, we have the following result, known as Rolle's theorem.

Theorem \(\PageIndex{2}\)

If \(f\) is continuous on \([a, b],\) differentiable on \((a, b),\) and \(f(a)=f(b),\) then there is a real number \(c\) in \((a, b)\) for which \(f^{\prime}(c)=0\).

More generally, suppose \(f\) is continuous on \([a, b]\) and differentiable on \((a, b) .\)

Let \[g(x)=f(x)-\frac{f(b)-f(a)}{b-a}(x-a)-f(a) .\] Note that \(g(x)\) is the difference between \(f(x)\) and the corresponding \(y\) value on the line passing through \((a, f(a))\) and \((b, f(b)) .\) Moreover, \(g\) is continuous on \([a, b],\) differentiable on \((a, b),\) and \(g(a)=0=g(b) .\) Hence Rolle's theorem applies to \(g,\) so there must exist a point \(c\) in \((a, b)\) for which \(g^{\prime}(c)=0 .\) Now \[g^{\prime}(c)=f^{\prime}(x)-\frac{f(b)-f(a)}{b-a} .\]

so we must have

\[0=g^{\prime}(c)=f^{\prime}(c)-\frac{f(b)-f(a)}{b-a} .\] That is, \[f^{\prime}(c)=\frac{f(b)-f(a)}{b-a} ,\] which is our desired connection between instantaneous and average rates of change, known as the mean-value theorem.

Theorem \(\PageIndex{3}\)

If \(f\) is continuous on \([a, b]\) and differentiable on \((a, b),\) then there exists a real number \(c\) in \((a, b)\) for which

\[f^{\prime}(c)=\frac{f(b)-f(a))}{b-a} .\]

Example \(\PageIndex{1}\)

Consider the function \(f(x)=x^{3}-3 x+1\) on the interval \([0,2]\). By the mean-value theorem, there must exist at least one point \(c\) in \([0,2]\) for which

\[f^{\prime}(c)=\frac{f(2)-f(0)}{2-0}=\frac{3-1}{2}=1 .\] Now \(f^{\prime}(x)=3 x^{2}-3,\) so \(f^{\prime}(c)=1\) implies \(3 c^{2}-3=1 .\) Hence \(c=\sqrt{\frac{4}{3}}\). Note that this implies that the tangent line to the graph of \(f\) at \(x=\sqrt{\frac{4}{3}}\) is parallel to the line through the endpoints of the graph of \(f,\) that is, the points \((0,1)\) and \((2,3) .\) See Figure \(1.9 .1 .\)

Increasing and Decreasing Functions

The preceding discussion leads us to the following definition and theorem.

Definition

We say a function \(f\) is increasing on an interval \(I\) if, whenever \(a<b\) are points in \(I, f(a)<f(b) .\) Similarly, we say \(f\) is decreasing on \(I\) if, whenever \(a<b\) are points in \(I, f(a)>f(b) .\)

Now suppose \(f\) is a defined on an interval \(I\) and \(f^{\prime}(x)>0\) for every \(x\) in \(I\) which is not an endpoint of \(I\). Then given any \(a\) and \(b\) in \(I,\) by the mean-value theorem there exists a point \(c\) between \(a\) and \(b\) for which

\[\frac{f(b)-f(a)}{b-a}=f^{\prime}(c)>0 .\] since \(b-a>0,\) this implies that \(f(b)>f(a) .\) Hence \(f\) is increasing on \(I .\) A similar argument shows that \(f\) is decreasing on \(I\) if \(f^{\prime}(x)<0\) for every \(x\) in \(I\) which is not an endpoint of \(I .\)

Theorem \(\PageIndex{4}\)

Suppose \(f\) is defined on an interval \(I .\) If \(f^{\prime}(x)>0\) for every \(x\) in \(I\) which is not an endpoint of \(I,\) then \(f\) is increasing on \(I .\) If \(f^{\prime}(x)<0\) for every \(x\) in \(I\) which is not an endpoint of \(I,\) then \(f\) is decreasing on \(I .\)

Example \(\PageIndex{2}\)

Let \(f(x)=2 x^{3}-3 x^{2}-12 x+1 .\) Then

\[f^{\prime}(x)=6 x^{2}-6 x-12=6\left(x^{2}-x-2\right)=6(x-2)(x+1) .\] Hence \(f^{\prime}(x)=0\) when \(x=-1\) and when \(x=2 .\) Now \(x-2<0\) for \(x<2\) and \(x-2>0\) for \(x>2,\) while \(x+1<0\) for \(x<-1\) and \(x+1>0\) when \(x>-1 .\) Thus \(f^{\prime}(x)>0\) when \(x<-1<\) and when \(x>2,\) and \(f^{\prime}(x)<0\) when \(-1<x<2 .\) It follows that \(f\) is increasing on the intervals \((-\infty,-1)\) and \((2, \infty),\) and decreasing on the interval \((-1,2) .\) Note that the theorem requires only that we know the sign of \(f^{\prime}\) at points inside a given interval, not at the endpoints. Hence it actually allows us to make the slightly stronger statement that \(f\) is increasing on the intervals \((-\infty,-1]\) and \([2, \infty),\) and decreasing on the interval \([-1,2] .\) Since \(f\) is increasing on \((-\infty,-1]\) and decreasing on \([-1,2],\) the point \((-1,8)\) must be a high point on the graph of \(f,\) although not necessarily the highest point on the graph. We say that \(f\) has a local maximum of 8 at \(x=-1\). Similarly, \(f\) is decreasing on \([-1,2]\) and increasing on \([2, \infty),\) and so the point \((2,-19)\) must be a low point on the graph of \(f,\) although, again, not necessarily the lowest point on the graph. We say that \(f\) has a local minimum of \(-19\) at \(x=2 .\) From this information, we can begin to see why the graph of \(f\) looks as it does in Figure \(1.9 .2 .\)

Definition

We say \(f\) has a local marimum at a point \(c\) if there exists an interval \((a, b)\) containing \(c\) for which \(f(c) \geq f(x)\) for all \(x\) in \((a, b) .\) Similarly, we say \(f\) has a local minimum at a point \(c\) if there exists an interval \((a, b)\) containing \(c\) for which \(f(c) \leq f(x)\) for all \(x\) in \((a, b) .\) We say \(f\) has a local extremum at \(c\) if \(f\) has either a local maximum or a local minimum at \(c .\)

We may now rephrase Theorem 1.9 .1 as follows.

Theorem \(\PageIndex{5}\)

If \(f\) is differentiable at \(c\) and has a local extremum at \(c,\) then \(f^{\prime}(c)=0\).

As illustrated in the preceding example, we may identify local minimums of a function \(f\) by locating those points at which \(f\) changes from decreasing to increasing, and local maximums by locating those points at which \(f\) changes from increasing to decreasing.

Example \(\PageIndex{3}\)

Let \(f(x)=x+2 \sin (x) .\) Then \(f^{\prime}(x)=1+2 \cos (x),\) and so \(f^{\prime}(x)<0\) when, and only when,

\[\cos (x)<-\frac{1}{2} .\] For \(0 \leq x \leq 2 \pi,\) this occurs when, and only when, \[\frac{2 \pi}{3}<x<\frac{4 \pi}{3} .\] Since the cosine function has period \(2 \pi,\) if follows that \(f^{\prime}(x)<0\) when, and only when, \(x\) is in an interval of the form \[\left(\frac{2 \pi}{3}+2 \pi n, \frac{4 \pi}{3}+2 \pi n\right)\] for \(n=0, \pm 1, \pm 2, \ldots\) Hence \(f\) is decreasing on these intervals and increasing on intervals of the form \[\left(-\frac{2 \pi}{3}+2 \pi n, \frac{2 \pi}{3}+2 \pi n\right) ,\]

\(n=0, \pm 1, \pm 2, \ldots\) It now follows that \(f\) has a local maximum at every point of the form

\[x=\frac{2 \pi}{3}+2 \pi n\] and a local minimum at every point of the form \[x=\frac{4 \pi}{3}+2 \pi n .\] From this information, we can begin to see why the graph of \(f\) looks as it does in Figure \(1.9 .3 .\)

Exercise \(\PageIndex{1}\)

Find the intervals where \(f(x)=x^{3}-6 x\) is increasing and the intervals where \(f\) is decreasing. Use this information to identify any local maximums or local minimums of \(f .\)

Answer

\(f\) is increasing on \((-\infty,-\sqrt{2})\) and on \([\sqrt{2}, \infty],\) decreasing on \([-\sqrt{2}, \sqrt{2}]\); \(f\) has a local maximum of \(4 \sqrt{2}\) at \(x=-\sqrt{2}\) and a local minimum of \(-4 \sqrt{2}\) at \(x=\sqrt{2} .\)

Exercise \(\PageIndex{2}\)

Find the intervals where \(f(x)=5 x^{3}-3 x^{5}\) is increasing and the intervals where \(f\) is decreasing. Use this information to identify any local maximums or local minimums of \(f .\)

Answer

\(f\) is increasing on \([-1,0]\) and on \([0,1]\) (or, simply, \([-1,1] ; f\) is decreasing on \((-\infty,-1]\) and on \([1, \infty) ; f\) has a local maximum of 2 at \(x=1\) and a local minimum of \(-2\) at \(x=-1 .\)

Exercise \(\PageIndex{3}\)

Find the intervals where \(f(x)=x+\sin (x)\) is increasing and the intervals where \(f\) is decreasing. Use this information to identify any local maximums or local minimums \(f .\)

Answer

\(f\) is increasing on all intervals of the form \([-n \pi, n \pi],\) where \(n=1,2, \dots ;\) \(f \text { has no local maximums or minimums (indeed, } f \text { is increasing on }(-\infty, \infty))\).