Higher order derivatives and graphs

$\newenvironment {prompt}{}{} \newcommand {\ungraded }[0]{} \newcommand {\todo }[0]{} \newcommand {\oiint }[0]{{\large \bigcirc }\kern -1.56em\iint } \newcommand {\mooculus }[0]{\textsf {\textbf {MOOC}\textnormal {\textsf {ULUS}}}} \newcommand {\npnoround }[0]{\nprounddigits {-1}} \newcommand {\npnoroundexp }[0]{\nproundexpdigits {-1}} \newcommand {\npunitcommand }[1]{\ensuremath {\mathrm {#1}}} \newcommand {\sagemath }[0]{\textsf {SageMath}} \newcommand {\surfaceColor }[0]{violet} \newcommand {\surfaceColorTwo }[0]{redyellow} \newcommand {\sliceColor }[0]{greenyellow} \newcommand {\vector }[1]{\left \langle #1\right \rangle } \newcommand {\sectionOutcomes }[0]{} \newcommand {\HyperFirstAtBeginDocument }[0]{\AtBeginDocument }$

Here we make a connection between a graph of a function and its derivative and higher order derivatives.

We say that a function $f$ is increasing on an interval $I$ if $f(x_{1})<f(x_{2})$ , for all pairs of numbers $x_{1}$ , $x_{2}$ in $I$ such that $x_{1}<x_{2}$ .
We say that a function $f$ is decreasing on an interval $I$ if $f(x_{1})>f(x_{2})$ , for all pairs of numbers $x_{1}$ , $x_{2}$ in $I$ such that $x_{1}<x_{2}$ .

Consider the graph of the function $f$ below:

On which of the following intervals is $f$ increasing?

$(-\infty ,a)$ $(-\infty ,b)$ $(a,b)$ $(a,+\infty )$ $(b,+\infty )$

The function $f$ is not increasing on the interval $(-\infty ,b)$ , because if we pick a pair of numbers from $(-\infty ,b)$ , say, $x_{1}=a$ , and $x_{2}=0$ , then $x_{1}<x_{2}$ , but $f(x_{1})>f(x_{2})$ .

Which of the following famous functions are increasing on $\Bigl (0,\frac {\pi }{2}\Bigr )$ ?

$\sin {(x)}$ $\sin {(2x)}$ $\cos {(x)}$ $\tan {(x)}$ $\cot {(x)}$ $f(x)=x^2$ $g(x)=\dfrac {1}{x}$

The function $\cos {(x)}$ is not increasing on $I=\Bigl (0,\frac {\pi }{2}\Bigr )$ , because if we take a pair of numbers in $I$ , say, $x_{1}=\frac {\pi }{6}$ , and $x_{2}=\frac {\pi }{3}$ , then $x_{1}<x_{2}$ , but
$f(x_{1})>f(x_{2})$ , since $f(x_{1})=\cos {\Bigl (\frac {\pi }{6}\Bigr )}=\frac {\sqrt {3}}{2}$ , and $f(x_{2})=\cos {}\Bigl (\frac {\pi }{3}\Bigr )=\frac {1}{2}$ .

Since the derivative gives us a formula for the slope of a tangent line to a curve, we can gain information about a function purely from the sign of the derivative. In particular, we have the following theorem

A function $f$ is increasing on any interval $I$ where $f'(x)>0$ , for all $x$ in $I$ .
A function $f$ is decreasing on any interval $I$ where $f'(x)<0$ , for all $x$ in $I$ .

Below we have graphed $y=f'(x)$ :

Is the function $f$ increasing or decreasing on the interval $-1<x<0$ ?

Increasing Decreasing

From the graph of $f'$ we can see that $f'(x)>0$ for all $x$ in $(-1,0)$ . Then, the Theorem above implies that the function $f$ is increasing on this interval.

We call the derivative of the derivative the second derivative, the derivative of the second derivative (the derivative of the derivative of the derivative) the third derivative, and so on. We have special notation for higher derivatives, check it out:

First derivative:: $\frac {d}{dx} f(x) = f'(x) = f^{(1)}(x)$ .
Second derivative:: $\frac {d}{dx^2} f(x) = f''(x) = f^{(2)}(x)$ .
Third derivative:: $\frac {d}{dx^3} f(x) = f'''(x) = f^{(3)}(x)$ .

We use the facts above in our next example.

Here we have unlabeled graphs of $f$ , $f'$ , and $f''$ :

Identify each curve above as a graph of $f$ , $f'$ , or $f''$ .

Here we see three curves, $A$ , $B$ , and $C$ . Since $A$ is positive negative increasing decreasing when $B$ is positive and positivenegativeincreasingdecreasing when $B$ is negative, we see $A'=B.$ Since $B$ is increasing when $C$ is positivenegativeincreasing decreasing and decreasing when $C$ is positivenegativeincreasingdecreasing , we see $B'=C.$ Hence $f=A$ , $f'=B$ , and $f''=C$ .

Here we have unlabeled graphs of $f$ , $f'$ , and $f''$ :

Identify each curve above as a graph of $f$ , $f'$ , or $f''$ .

Here we see three curves, $A$ , $B$ , and $C$ . Since $B$ is positivenegativeincreasingdecreasing when $A$ is positive and positivenegativeincreasingdecreasing when $A$ is negative, we see $B'=A.$ Since $A$ is increasing when $C$ is positivenegativeincreasingdecreasing and decreasing when $C$ is positivenegativeincreasingdecreasing , we see $A'=C.$ Hence $f=\answer [given]{B}$ , $f'=\answer [given]{A}$ , and $f''=\answer [given]{C}$ .

Here we have unlabeled graphs of $f$ , $f'$ , and $f''$ :

Identify each curve above as a graph of $f$ , $f'$ , or $f''$ .

Here we see three curves, $A$ , $B$ , and $C$ . Since $C$ is positivenegativeincreasingdecreasing when $B$ is positive and positivenegativeincreasingdecreasing when $B$ is negative, we see $C'=B.$ Since $B$ is increasing when $A$ is positivenegativeincreasingdecreasing and decreasing when $A$ is positivenegativeincreasingdecreasing , we see $B'=A.$ Hence $f=\answer [given]{C}$ , $f'=\answer [given]{B}$ , and $f''=\answer [given]{A}$ .