Extrema

A local extremum of a function $f$ is a point $(a,f(a))$ on the graph of $f$ where the $y$-coordinate is larger (or smaller) than all other $y$-coordinates of points on the graph whose $x$-coordinates are “close to” $a$.

In our next example, we clarify the definition of a local minimum.

Consider the graph of a function $f$:

Identify the local extrema of $f$ and give an explanation.

In the figure above the function $f$ has a local minimum at, $$a=\answer [given]{1},$$ because we can find an open interval $I$ (marked in the figure) that contains the number $$a=\answer [given]{1},$$ and for all numbers $x$ in $I$ the following statement is true: $$f\left (\answer [given]{1}\right )\le f\left (\answer [given]{x}\right ).$$

Local maximum and minimum points are quite distinctive on the graph of a function, and are, therefore, useful in understanding the shape of the graph. Many problems in real world and in different scientific fields turn out to be about finding the smallest (or largest) value that a function achieves (for example, we might want to find the minimum cost at which some task can be performed).

Critical points

Consider the graph of the function $f$.

After this example, the following theorem should not come as a surprise.

Fermat’s Theorem says that the only points at which a function can have a local maximum or minimum are points at which the derivative is zero or the derivative is undefined. As an illustration of the first scenario, consider the plots of $f(x) = x^3-4.5x^2+6x$ and $f'(x) = 3x^2-9x+6$.

This brings us to our next definition.

Since both local maximum and local minimum occur at a critical point, when we locate a critical point, we need to determine which, if either, actually occurs.

Find all local maximum and minimum points for the function $f(x)=x^3-x$.

Write $$\ddx f(x)=\answer [given]{3x^2-1}.$$ We can easily express $f'(x)$ as a product of its factors $$\ddx f(x)=3\left (x+\answer [given]{\frac {\sqrt {3}}{3}}\right )\left (x-\answer [given]{\frac {\sqrt {3}}{3}}\right ),$$ which implies that the function $f$ has only two critical points, $x=-\frac {\sqrt {3}}{3}$ and $x=\frac {\sqrt {3}}{3}$. Notice that the derivative $f'(x)$ is a polynomial, and polynomials do not change signs except possibly at their zeros. This implies that derivative $f'(x)$ does not change the sign on the intervals $\left (-\infty ,-\frac {\sqrt {3}}{3}\right )$, $\left (-\frac {\sqrt {3}}{3},\frac {\sqrt {3}}{3}\right )$, and $\left (\frac {\sqrt {3}}{3},\infty \right )$, because these intervals do not contain any zeros of $f'(x)$.

Select the correct statement about the sign of $f'(x)$ on the intervals $\left (-\infty ,-\frac {\sqrt {3}}{3}\right )$ and $\left (-\frac {\sqrt {3}}{3},\frac {\sqrt {3}}{3}\right )$.

$f'(x)>0$ on $\left (-\infty ,-\frac {\sqrt {3}}{3}\right )$ and $f'(x)>0$ on $\left (-\frac {\sqrt {3}}{3},\frac {\sqrt {3}}{3}\right )$. $f'(x)>0$ on $\left (-\infty ,-\frac {\sqrt {3}}{3}\right )$ and $f'(x)<0$ on $\left (-\frac {\sqrt {3}}{3},\frac {\sqrt {3}}{3}\right )$. $f'(x)<0$ on $\left (-\infty ,-\frac {\sqrt {3}}{3}\right )$ and $f'(x)>0$ on $\left (-\frac {\sqrt {3}}{3},\frac {\sqrt {3}}{3}\right )$. $f'(x)<0$ on $\left (-\infty ,-\frac {\sqrt {3}}{3}\right )$ and $f'(x)<0$ on $\left (-\frac {\sqrt {3}}{3},\frac {\sqrt {3}}{3}\right )$

If we know the sign of the derivative on an interval, we also know whether the function is increasing or decreasing on that interval. This will help us determine whether the function has a local extremum at the critical point where $x=-\frac {\sqrt {3}}{3}$.

At the critical point where $x=-\frac {\sqrt {3}}{3}$, the function $f$ has

no local extremum, because $f$ is increasing on $\left (-\infty ,-\frac {\sqrt {3}}{3}\right )$ and increasing on $\left (-\frac {\sqrt {3}}{3},\frac {\sqrt {3}}{3}\right )$. a local maximum, because $f$ is increasing on $\left (-\infty ,-\frac {\sqrt {3}}{3}\right )$ and decreasing on $\left (-\frac {\sqrt {3}}{3},\frac {\sqrt {3}}{3}\right )$. a local minimum, because $f$ is decreasing on $\left (-\infty ,-\frac {\sqrt {3}}{3}\right )$ and increasing on $\left (-\frac {\sqrt {3}}{3},\frac {\sqrt {3}}{3}\right )$. no local extremum, because $f$ is decreasing on $\left (-\infty ,-\frac {\sqrt {3}}{3}\right )$ and decreasing on $\left (-\frac {\sqrt {3}}{3},\frac {\sqrt {3}}{3}\right )$.

Select the correct statement about the sign of $f'(x)$ on the intervals $\left (-\frac {\sqrt {3}}{3},\frac {\sqrt {3}}{3}\right )$ and $\left (\frac {\sqrt {3}}{3},\infty \right )$.

$f'(x)>0$ on $\left (-\frac {\sqrt {3}}{3},\frac {\sqrt {3}}{3}\right )$ and $f'(x)>0$ on $\left (\frac {\sqrt {3}}{3},\infty \right )$. $f'(x)>0$ on $\left (-\frac {\sqrt {3}}{3},\frac {\sqrt {3}}{3}\right )$ and $f'(x)<0$ on $\left (\frac {\sqrt {3}}{3},\infty \right )$. $f'(x)<0$ on $\left (-\frac {\sqrt {3}}{3},\frac {\sqrt {3}}{3}\right )$ and $f'(x)>0$ on $\left (\frac {\sqrt {3}}{3},\infty \right )$. $f'(x)<0$ on $\left (-\frac {\sqrt {3}}{3},\frac {\sqrt {3}}{3}\right )$ and $f'(x)<0$ on $\left (\frac {\sqrt {3}}{3},\infty \right )$.

Again, the sign of the derivative on an interval determines whether the function is increasing or decreasing on that interval. This will help us determine whether the function has a local extremum at the critical point where $x=\frac {\sqrt {3}}{3}$.

At the critical point where $x=\frac {\sqrt {3}}{3}$, the function $f$ has

no local extremum, because $f$ is increasing on $\left (-\frac {\sqrt {3}}{3},\frac {\sqrt {3}}{3}\right )$ and increasing on $\left (\frac {\sqrt {3}}{3},\infty \right )$. a local maximum, because $f$ is increasing on $\left (-\frac {\sqrt {3}}{3},\frac {\sqrt {3}}{3}\right )$ and decreasing on $\left (\frac {\sqrt {3}}{3},\infty \right )$. a local minimum, because $f$ is decreasing on $\left (-\frac {\sqrt {3}}{3},\frac {\sqrt {3}}{3}\right )$ and increasing on $\left (\frac {\sqrt {3}}{3},\infty \right )$. no local extremum, because $f$ is decreasing on$\left (-\frac {\sqrt {3}}{3},\frac {\sqrt {3}}{3}\right )$ and decreasing on $\left (\frac {\sqrt {3}}{3},\infty \right )$.

Do your answers agree with the graphs of $f$ and $f'$ given in the picture below?

The first derivative test

We will further explore and refine the method of the previous section for deciding whether there is a local maximum or minimum at a critical point. Recall that

• If $f'(x) >0$ on an interval, then $f$ is increasing on that interval.
• If $f'(x) <0$ on an interval, then $f$ is decreasing on that interval.

So how exactly does the derivative tell us whether there is a maximum, minimum, or neither at a point? Use the first derivative test.

Consider the function $$f(x) = \frac {x^4}{4}+\frac {x^3}{3}-x^2$$ Find the intervals on which $f$ is increasing, the intervals on which $f$ is decreasing and identify the local extrema of $f$.

Start by computing $$\ddx f(x) = \answer [given]{x^3+x^2-2x}.$$ Now we need to find on what intervals is $f'$ positive and what intervals it is negative. To do this, solve $$f'(x) = \answer [given]{x^3+x^2-2x} =0.$$ Factor $f'(x)$ $$\begin{align*} f'(x) &= \answer[given]{x^3+x^2-2x} \\ &=x(\answer[given]{x^2+x-2})\\ &=x(x+2)(\answer[given]{x-1}). \end{align*}$$ So the critical points (when $f'(x)=0$) are when $x=-2$, $x=0$, and $x=1$. Since the derivative, $f'(x)$, is a polynomial, it does not change the sign on intervals between its zeros, i.e., between the critical points. Now we can check the sign of $f'(x)$ at some points between the critical points to find where $f'(x)$ is positive and where negative: $$\begin{align*} f'(-3)&=\answer[given]{-12},\\ f'(0.5)&=\answer[given]{-0.625},\\ f'(-1)&=\answer[given]{2},\\ f'(2)&=\answer[given]{8}. \end{align*}$$ From this we can make a sign table:

Hence $f$ is increasing on $(-2,0)$ and $(1,\infty )$ and $f$ is decreasing on $(-\infty ,-2)$ and $(0,1)$. Moreover, from the first derivative test, the local maximum is at $x=0$ while the local minimums are at $x=-2$ and $x=1$.

This can be confirmed by checking the graphs of $f(x) =x^4/4 + x^3/3 -x^2$ and $f'(x) = x^3 + x^2 -2x$.

Hence we have seen that if $f'$ is zero at a point and increasing on an interval containing that point, then $f$ has a local minimum at the point. If $f'$ is zero at a point and decreasing on an interval containing that point, then $f$ has a local maximum at the point. Thus, we see that we can gain information about $f$ by studying how $f'$ changes. This leads us to our next section.

Inflection points

If we are trying to understand the shape of the graph of a function, knowing where it is concave up and concave down helps us get a more accurate picture. It is worth summarizing what we have already seen into a single theorem.

Of particular interest are points at which the concavity changes from up to down or down to up.

It is instructive to see some examples of inflection points:

It is also instructive to see some nonexamples of inflection points:

We identify inflection points by first finding $x$ such that $f''(x)$ is zero or undefined and then checking to see whether $f''(x)$ does in fact go from positive to negative or negative to positive at these points.

Describe the concavity of $f(x)=x^3-x$.

To start, compute the first and second derivative of $f(x)$ with respect to $x$, $$f'(x)=\answer [given]{3x^2-1}\qquad \text {and}\qquad f''(x)=\answer [given]{6x}.$$ Since $f''(0)=0$, there is potentially an inflection point at $x=0$. Using test points, we note the concavity does change from down to up, hence there is an inflection point at $x=0$. The curve is concave down for all $x<0$ and concave up for all $x>0$, see the graphs of $f(x) = x^3-x$ and $f''(x) = 6x$.

Note that we need to compute and analyze the second derivative to understand concavity, so we may as well try to use the second derivative test for maxima and minima. If for some reason this fails we can then try one of the other tests.

The second derivative test

Recall the first derivative test:

• If $f'(x)>0$ for all $x$ that are near and to the left of $a$ and $f'(x)<0$ for all $x$ that are near and to the right of $a$, then the function $f$ has a local maximum at $a$.
• If $f'(x)<0$ for all $x$ that are near and to the left of $a$ and $f'(x)>0$ for all $x$ that are near and to the right of $a$, then the function $f$ has a local minimum at $a$.

Assume that a function $f$ has a critical point at $a$ and that $f'(a)=0$. If $f'$ happens to be decreasing on some interval containing $a$, then it changes from positive to negative at $a$. Therefore, if $f''$ is negative on some interval that contains $a$, then $f'$ is definitely decreasing, so there is a local maximum at the point in question. On the other hand, if $f'$ is increasing, then it changes from negative to positive at $a$. Therefore, if $f''$ is positive on an interval that contains $a$, then $f'$ is definitely increasing, so there is a local minimum at the point in question. We summarize this as the second derivative test.

The second derivative test is often the easiest way to identify local maximum and minimum points. Sometimes the test fails and sometimes the second derivative is quite difficult to evaluate. In such cases we must fall back on one of the previous tests.

Once again, consider the function $$f(x) = \frac {x^4}{4}+\frac {x^3}{3}-x^2$$ Use the second derivative test to locate the local extrema of $f$.

Start by computing $$f'(x) = \answer [given]{x^3 + x^2 -2x} \qquad \text {and}\qquad f''(x) = \answer [given]{3x^2 + 2x-2}.$$ Using the same technique as we used before, we find that $$f'(-2) = \answer [given]{0},\qquad f'(0) = \answer [given]{0}, \qquad f'(1) = \answer [given]{0}.$$ Now we’ll attempt to use the second derivative test, $$f''(-2) = \answer [given]{6}, \qquad f''(0) =\answer [given]{ -2}, \qquad f''(1) = \answer [given]{3}.$$ Hence we see that $f$ has a local minimum at $x=-2$, a local maximum at $x=0$, and a local minimum at $x=1$, see below for a plot of $f(x) =x^4/4 + x^3/3 -x^2$ and $f''(x) = 3x^2 + 2x -2$: