The First Derivative Test

The derivative $f'(x)$ can tell us a lot about the function $y=f(x)$. It can tell us where critical points are i.e. points at which $f'(x)=0$ and the critical points are likely places at which $y=f(x)$ assumes a local maximum or a local minimum values. By further examining the properties of $f'(x)$ we can also determine at which critical point, $f(x)$ assumes a local maximum, or a local minimum, or neither. But first we see that $f'(x)$ can tell us where $y=f(x)$ is increasing or decreasing.

Theorem. Increasing/Decreasing Test

1. If $f'(x)>0$ on an open interval, $f$ is increasing on that interval.
2. If $f'(x)<0$ on an open interval, $f$ is decreasing on that interval.

Example. Find where $f(x)=3x^4-4x^3-12x^2+5$ is increasing and where it is decreasing.

Solution.
\begin{align*}
f'(x)&=12x^3-12x^2-24x\\
&=12x(x^2-x-2)\\
&=12x(x-2)(x+1).
\end{align*}
The critical points are $x=-1,0,2$. Using, for instance, the test point method (which is the easiest method of solving an inequality), we obtain the following table.
$$
\begin{array}{|c|c|c|c|c|c|c|c|}
\hline
x & x<-1 & -1 & -1<x<0 & 0 & 0<x<2 & 2 & x>2\\
\hline
f'(x) & – & 0 & + & 0 & – & 0 & +\\
\hline
f(x) & \searrow & f(-1) & \nearrow & f(0) & \searrow & f(2) &\nearrow\\
\hline
\end{array}
$$
So we find that $f$ is increasing on $(-1,0)\cup(2,\infty)$ and $f$ is decreasing on $(-\infty,-1)\cup(0,2)$.

Now, local maximum values and local minimum values can be identified by observing the change of sign of $f'(x)$ at each critical point.

Theorem. [The First Derivative Test] Suppose that $c$ is a critical point of a differentiable function $f(x)$.

1. If the sign of $f'(x)$ changes from $+$ to $-$ at $c$, $f(c)$ is a local maximum.
2. If the sign of $f'(x)$ changes from $-$ to $+$ at $c$, $f(c)$ is a local minimum.
3. If the sign $f'(x)$ does not change at $c$, $f$ has neither a local maximum nor a local minimum at $c$.

Example. In the previous example, the sign of $f'(x)$ changes from $+$ to $-$ at $0$, so $f(0)=5$ is a local maximum. The sign of $f'(x)$ changes from $-$ to $+$ at $-1$ and at $2$, so $f(-1)=0$ and $f(2)=-27$ are local minimum values.

The following figure confirms our findings from the above two examples.

The Second Derivative Test

The second order derivative $f^{\prime\prime}(x)$ can provide us an additional piece of information on $y=f(x)$, namely the concavity of the graph of $y=f(x)$.

Definition. If the graph of $f$ lies above all of its tangents on an open interval $I$, it is called concave upward on $I$. If the graph of $f$ lies below all of its tangents on $I$, it is called concave downward on $I$.

From here on, $\smile$ denotes “concave up” and $\frown$ denotes “concave down”.

Definition. A point $(d,f(d))$ on the graph of $y=f(x)$ is called a point of inflection if the concavity of the graph of $f$ changes from $\smile$ to $\frown$ or from $\frown$ to $\smile$ at $(d,f(d))$. The candidates for the points of inflection may be found by solving the equation $f^{\prime\prime}(x)=0$ as shown in the example below.

Theorem. [Concavity Test]

1. If $f^{\prime\prime}(x)>0$ for all x in an open interval $I$, the graph of $f$ is concave up on $I$.
2. If $f^{\prime\prime}(x)<0$ for all x in an open interval $I$, the graph of $f$ is concave down on $I$.

Theorem. [The Second Derivative Test] Suppose that $f'(c)=0$ i.e. $c$ is a critical point of $f$. Suppose that $f^{\prime\prime}$ is continuous near $c$.

1. If $f^{\prime\prime}(c)>0$ then $f(c)$ is a local minimum.
2. If $f^{\prime\prime}(c)<0$ then $f(c)$ is a local maximum.

Example. Let $f(x)=-x^4+2x^2+2$.

1. Find and identify all local maximum and local minimum values of $f(x)$ using the Second Derivative Test.
2. Find the intervals on which the graph of $f(x)$ is concave up or concave down. Find all points of inflection.

Solution. 1. First we find the critical points of $f(x)$ by solving the equation $f'(x)=0$:
$$f'(x)=-4x^3+4x=-4x(x^2-1)=-4x(x+1)(x-1)=0.$$ So $x=-1,0,1$ are critical points of $f(x)$ Next, $f^{\prime\prime}(x)=-12x^2+4$. Since $f^{\prime\prime}(0)=4>0$ and $f^{\prime\prime}(-1)=f^{\prime\prime}(1)=-8<0$, by the Second Derivative Test, $f(0)=2$ is a local minimum value and $f(-1)=f(1)=3$ is a local maximum value.

2. First we need to solve the equation $f”(x)=0$:
$$f^{\prime\prime}(x)=-12x^2+4=-12\left(x^2-\frac{1}{3}\right)=-12\left(x+\frac{1}{\sqrt{3}}\right)\left(x-\frac{1}{\sqrt{3}}\right)=0.$$ So $f^{\prime\prime}(x)=0$ at $x=\pm\displaystyle\frac{1}{\sqrt{3}}$. By using the test-point method we find the following table:
$$
\begin{array}{|c||c|c|c|c|c|}
\hline
x & x<-\frac{1}{\sqrt{3}} & -\frac{1}{\sqrt{3}} & -\frac{1}{\sqrt{3}}<x<\frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} & x>\frac{1}{\sqrt{3}}\\
\hline
f^{\prime\prime}(x) & – & 0 & + & 0 & -\\
\hline
f(x) & \frown & f\left(-\frac{1}{\sqrt{3}}\right)=\frac{23}{9} & \smile & f\left(\frac{1}{\sqrt{3}}\right)=\frac{23}{9} & \frown\\
\hline
\end{array}
$$
The graph of $f(x)$ is concave down on the intervals $\left(-\infty,-\frac{1}{\sqrt{3}}\right)\cup\left(\frac{1}{\sqrt{3}},\infty\right)$ and is concave up on the interval $\left(-\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}\right)$. The points of inflection are $\left(-\frac{1}{\sqrt{3}},\frac{23}{9}\right)$ and $\left(\frac{1}{\sqrt{3}},\frac{23}{9}\right)$.

The following figure confirms our findings from the above example.

If given integration takes the form $\int f(g(x))g'(x)dx$ then it can be converted to a simpler integration that we may be able to evaluate by the substitution $u=g(x)$. In fact, the integration is given in terms of the new variable $u$ as
$$\int f(g(x))g'(x)dx=\int f(u)du.$$

Example. Evaluate $\int x\sqrt{1+x^2}dx$.

Solution. Let $u=1+x^2$. Then $du=2xdx$. So,
\begin{align*}
\int x\sqrt{1+x^2}dx&=\frac{1}{2}\int\sqrt{u}du\\
&=\frac{1}{3}u^{\frac{3}{2}}+C\\
&=\frac{1}{3}(1+x^2)^{\frac{3}{2}}+C,
\end{align*}
where $C$ is an arbitrary constant.

Example. Evaluate $\int\cos (7\theta+5)d\theta$.

Solution. Let $u=7\theta+5$. Then $du=7d\theta$. So,
\begin{align*}
\int\cos (7\theta+5)d\theta&=\frac{1}{7}\int\cos udu\\
&=\frac{1}{7}\sin u+C\\
&=\frac{1}{7}\sin(7\theta+5)+C,
\end{align*}
where $C$ is an arbitrary constant.

Example. Evaluate $\int x^2\sin(x^3)dx$.

Solution. Let $u=x^3$. Then $du=3x^2dx$. So,
\begin{align*}
\int x^2\sin(x^3)dx&=\frac{1}{3}\int \sin udu\\
&=-\frac{1}{3}\cos u+C\\
&=-\frac{1}{3}\cos(x^3)+C,
\end{align*}
where $C$ is an arbitrary constant.

Example. Integrals of $\sin^2x$ and $\cos^2x$.

1. \begin{align*}\int\sin^2xdx&=\int\frac{1-\cos 2x}{2}dx\\&=\frac{x}{2}-\frac{\sin 2x}{4}+C\end{align*}
2. \begin{align*}\int\cos^2xdx&=\int\frac{1+\cos 2x}{2}dx\\&=\frac{x}{2}+\frac{\sin 2x}{4}+C\end{align*}

How do we evaluate a definite integral of the form $\int_a^b f(g(x))g'(x)dx$? The following example shows you how.

Example. Evaluate $\int_{-1}^1 3x^2\sqrt{x^3+1}dx$.

Solution. First let us calculate the indefinite integral $\int 3x^2\sqrt{x^3+1}dx$. Let $u=x^3+1$. Then $du=3x^2dx$. So,
\begin{align*}
\int 3x^2\sqrt{x^3+1}dx&=\int\sqrt{u}du\\
&=\frac{2}{3}u^{\frac{3}{2}}+C\\
&=2(x^3+1)^{\frac{3}{2}}+C,
\end{align*}
where $C$ is an arbitrary constant. Now by Fundamental Theorem of Calculus,
\begin{align*}
\int_{-1}^1 3x^2\sqrt{x^3+1}dx&=\frac{2}{3}[(x^3+1)^{\frac{3}{2}}]_{-1}^1\\
&=\frac{4\sqrt{2}}{3}.
\end{align*}

But there is a better way to do this as shown in the following theorem. Its proof is straightforward.

Theorem. If $u’$ is continuous on $[a,b]$ and $f$ is continuous on the range of $u$, then
$$\int_a^bf(u(x))u'(x)dx=\int_{u(a)}^{u(b)}f(u)du.$$

Example. Let us replay the previous example using this theorem. Again let $u=x^3+1$. Then $du=3x^2dx$ and $u(-1)=0$, $u(1)=2$. Now, by the above theorem,
\begin{align*}
\int_{-1}^1 3x^2\sqrt{x^3+1}dx&=\int_0^2\sqrt{u}du\\
&=\frac{2}{3}[u^{\frac{3}{2}}]_0^2\\
&=\frac{4\sqrt{2}}{3}.
\end{align*}
I believe you will find this more simple than previous method.

Example. Evaluate $\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\cot\theta\csc^2\theta d\theta$.

Solution. Let $u=\cot\theta$. Then $u\left(\frac{\pi}{4}\right)=1$, $u\left(\frac{\pi}{2}\right)=0$, and $du=-\csc^2\theta d\theta$. Hence, \begin{align*}\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\cot\theta\csc^2\theta d\theta&=-\int_1^0 udu\\&=\frac{1}{2}\end{align*}

I will finish this lecture with the following nice properties.

Theorem. Let $f$ be a continuous function on $[-a,a]$.

(a) If $f$ is an even function, then $\int_{-a}^a f(x)dx=2\int_0^a f(x)dx$.

(b) If $f$ is an odd function, then $\int_{-a}^a f(x)dx=0$.

This can be easily understood from pictures using the symmetries of even and odd functions. But the theorem can be proved using substitution. I will leave it to you.

Example.  \begin{align*}\int_{-2}^2(x^4-4x^2+6)dx&=2\int_0^2(x^4-4x^2+6)dx\\&=2\left[\frac{x^5}{5}-\frac{4}{3}x^3+6x\right]_0^2\\&=\frac{232}{15}\end{align*}

The following theorem is something one can easily picture intuitively.

Theorem. [Rolle’s Theorem]
Let $f$ be continuous on the closed interval $[a,b]$ and differentiable on the open interval $(a,b)$. If $f(a)=f(b)$, then there exists a number $c$ in $(a,b)$ such that $f'(c)=0$.

Example. Show that the equation $x^3+x-1=0$ has exactly only one real root.

Solution. Let $f(x)=x^3+x-1$. Note that $f(0)=-1$ and $f(1)=1$. So by the Intermediate Value Theorem, we see that there exists at least a root of the equation $x^3+x-1=0$ in the interval $(0,1)$. Now suppose that there are two different roots $a$ and $b$ of the equation $x^3+x-1=0$. Without loss of generality, we may assume that $a<b$. Then $f(x)$ is continuous on $[a,b]$ and differentiable on $(a,b)$. By Rolle’s Theorem then, there exist a number $c$ in $(a,b)$ such that $f'(c)=0$. However, $f'(x)=3x^2+1\geq 1$ for all real number $x$. This is a contradiction. Therefore, there should be only one root of the equation.

Let $f$ be continuous on $[a,b]$ and differentiable on $(a,b)$. Define $g(x)$ to be the distance between $f(x)$ and the line segment from $(a,f(a))$ to $(b,f(b))$, i.e.
$$g(x)=f(x)-\frac{f(b)-f(a)}{b-a}(x-a)-f(a).$$ Then $g(x)$ is continuous on $[a,b]$ and differentiable on $(a,b)$. Since $g(a)=g(b)=0$, by Rolle’s theorem there exists a number $c$ in $(a,b)$ such that $g'(c)=f'(c)-\frac{f(b)-f(a)}{b-a}=0$. Therefore, we proved the following theorem.

Theorem. [Mean Value Theorem]
Let $f$ be continuous on $[a,b]$ and differentiable on $(a,b)$. Then there exists a number $c$ in $(a,b)$ such that
$$f'(c)=\frac{f(b)-f(a)}{b-a}.$$

Remark. Why the name Mean Value Theorem? The average $\mathrm{av}(f)$ of a continuous function $f$ on a closed interval $[a,b]$ can be defined by $$\mathrm{av}(f)=\frac{1}{b-a}\int_a^b f(x)dx$$ See here for details. If $f'(x)$ is continuous on $[a,b]$, its average on $[a,b]$ is given by $$\frac{1}{b-a}\int_a^bf'(x)dx=\frac{f(b)-f(a)}{b-a}$$ That is, Mean Value Theorem states that one of the values of $f'(x)$ on $(a,b)$ becomes the average of $f'(x)$ on $[a,b]$.

The following examples are applications of the Mean Value Theorem.

Example. Suppose that $f(0)=-3$ and $f'(x)\leq 5$ for all values of $x$. How large can $f(2)$ possibly be?

Solution. By the Mean Value Theorem, there exists a number $c$ in $(0,2)$ such that
$$f'(c)=\frac{f(2)-f(0)}{2-0}=\frac{f(2)+3}{2}.$$
Since $f'(c)\leq 5$,
\begin{align*}
f(2)&=2f'(c)-3\\
&\leq 2\cdot 5-3=7.
\end{align*}
Hence, $7$ is the largest possible value of $f(2)$.

Example. A trucker handed in a ticket at a toll booth showing in 2 hours she had covered 159 mi on a toll road with speed limit of 65 mph. The trucker was cited for speeding. Why?

Solution. The average speed was $\frac{159}{2}=79.5$ mph. By the Mean Value Theorem the trucker was driving at the speed 79.5 mph at some point .

Using Mean Value Theorem, one can prove the following theorem. The proof is straightforward and is left for readers.

Theorem. If $f'(x)=0$ for all $x$ in the open interval $(a,b)$, then $f$ is constant on $(a,b)$.

Maximum and Minimum

There are two different types of extremum (maximum or minimum) values of a function $y=f(x)$. We may consider a value of $y$ that is an extremum globally on the domain or we may also consider a value of $y$ that is an extremum locally around an $x$ value.

A function $f$ has an absolute maximum at $c$ if $f(c)\geq f(x)$ for all $x$ in the domain of $f$. Similarly, $f$ has an absolute minimum at $c$ if $f(c)\leq f(x)$ for all $x$ in the domain of $f$.

A function $f$ has a local maximum (or relative maximum) at $c$ if $f(c)\geq f(x)$ in some neighborhood of $c$ (i.e an open interval that contains $c$). Similarly, $f$ has a local minimum (or relative minimum) at $c$ if $f(c)\leq f(x)$ in some neighborhood of $c$.

Example.

The above figure shows the graph of $f(x)=3x^4-16x^3+18x^2$, $-1\leq x\leq 4$. It has a local maximum at $x=1$ and a local minimum at $x=3$. The local minimum $f(3)=-27$ is also an absolute minimum. $f$ has an absolute maximum $f(-1)=37$. This $f(-1)=37$ is not a local maximum by the way. The reason is that there is no local neighborhood around $x=-1$ as the domain is given by $[-1,4]$.

A natural question one may ask is whether a function always has an absolute maximum and an absolute minimum. You can easily find many examples that show that a function does not necessarily have an absolute maximum or an absolute minimum value. For instance, $y=x$ on $(-\infty,\infty)$ has neither an absolute maximum nor an absolute minimum. The function $y=x^2$ on $[0,1)$ has an absolute minimum 0 at $x=0$ but has no absolute maximum.

Theorem. [Max-Min Theorem, Fermat]
If $f$ is continuous on a closed interval $[a,b]$, then $f$ attains an absolute maximum and an absolute minimum on $[a,b]$.

The following theorem is also due to Fermat.

Theorem. If $f$ has a local maximum or a local minimum at $c$ and if $f'(c)$ exists, then $f'(c)=0$.

The converse of this theorem is not necessarily true i.e. $f'(c)=0$ does not necessarily mean that $f(c)$ is a local maximum or a local minimum. For example, consider $f(x)=x^3$. $f'(0)=0$ but $f(x)$ has neither a local maximum nor a local minimum at $x=0$ as shown in figure below.

The above theorem is important as an absolute maximum and an absolute minimum may be found among local maximum values, local minimum values and the evaluations of $f$ at the end points, $f(a)$ and $f(b)$. To find local maximum values and local minimum values, we first find points $c$ such that $f'(c)=0$. Such points are called critical points. The reason they are called critical points is that the graph of a function changes from increasing to decreasing or from decreasing to increasing at a critical point.

Definition. A critical point of a function $f(x)$ is a number $c$ in the domain of $f$ such that either $f'(c)=0$ or $f'(c)$ does not exist.

Recipe of Finding Absolute Maximum and Absolute Minimum

Let $f$ be a continuous function on a closed interval $[a,b]$.

Step 1. Find all critical points of $f$ in $(a,b)$.

Step 2. Evaluate $f$ at each critical point obtained in Step 1.

Step 3. Find $f(a)$ and $f(b)$.

Step 4. Compare all the values obtained in Steps 2 and 3. The largest value is the absolute maximum and the smallest value is the absolute minimum.

Example. Find the absolute maximum and the absolute minimum values of
$$f(x)=x^3-3x^2+1,\ -\frac{1}{2}\leq x\leq 4.$$

Solution.

Step 1. Find all critical points of $f$ in $\left(-\frac{1}{2},4\right)$.

$f'(x)=3x^2-6x$. Set $f'(x)=0$ i.e. $3x^2-6x=0$. $3x^2-6x$ is factored as $3x(x-2)$. So we find two critical points $0, 2$.

Step 2. Evaluate $f$ at each critical point obtained in Step 1.

$f(0)=1$ and $f(2)=-3$.

Step 3. Find $f\left(-\frac{1}{2}\right)$ and $f(4)$.

$f\left(-\frac{1}{2}\right)=\frac{1}{8}$ and $f(4)=17$.

Step 4. Compare all the values obtained in Steps 2 and 3.

The largest value is $f(4)=17$ so this is the absolute maximum value of $f$ on $\left[-\frac{1}{2},4\right]$. The smallest value is $f(2)=-3$ so this is the absolute minimum of $f$ on $\left[-\frac{1}{2},4\right]$.