3.1 Critical Numbers

$\newenvironment {prompt}{}{} \newcommand {\ungraded }[0]{} \newcommand {\ffrac }[2]{\frac {\text {\footnotesize $#1$}}{\text {\footnotesize $#2$}}} \newcommand {\newrobustcmd }[0]{} \newcommand {\csshow }[1]{\begingroup \expandafter \endgroup \expandafter \show \csname #1\endcsname } \newcommand {\csmeaning }[1]{\ifcsname #1\endcsname \expandafter \meaning \csname #1\endcsname \else \detokenize {undefined}\fi } \newcommand {\ifdefmacro }[0]{} \newcommand {\ifdefparam }[0]{} \newcommand {\ifdefprotected }[0]{} \newcommand {\ifnumequal }[1]{\ifnumcomp {#1}=} \newcommand {\ifnumgreater }[1]{\ifnumcomp {#1}>} \newcommand {\ifnumless }[1]{\ifnumcomp {#1}<} \newcommand {\ifdimequal }[1]{\ifdimcomp {#1}=} \newcommand {\ifdimgreater }[1]{\ifdimcomp {#1}>} \newcommand {\ifdimless }[1]{\ifdimcomp {#1}<} \newcommand {\expandonce }[1]{\unexpanded \expandafter {#1}} \newcommand {\csexpandonce }[1]{\expandafter \expandonce \csname #1\endcsname } \newcommand {\protecting }[0]{} \newcommand {\csdef }[1]{\expandafter \def \csname #1\endcsname } \newcommand {\csedef }[1]{\expandafter \edef \csname #1\endcsname } \newcommand {\csgdef }[1]{\expandafter \gdef \csname #1\endcsname } \newcommand {\csxdef }[1]{\expandafter \xdef \csname #1\endcsname } \newcommand {\cslet }[2]{\expandafter \let \csname #1\endcsname #2} \newcommand {\letcs }[2]{\ifcsdef {#2} {\expandafter \let \expandafter #1\csname #2\endcsname } {\undef #1}} \newcommand {\csletcs }[2]{\ifcsdef {#2} {\expandafter \let \csname #1\expandafter \endcsname \csname #2\endcsname } {\csundef {#1}}} \newcommand {\csuse }[1]{\ifcsname #1\endcsname \csname #1\expandafter \endcsname \fi } \newcommand {\appto }[2]{\ifundef {#1} {\edef #1{\unexpanded {#2}}} {\edef #1{\expandonce #1\unexpanded {#2}}}} \newcommand {\eappto }[2]{\ifundef {#1} {\edef #1{#2}} {\edef #1{\expandonce #1#2}}} \newcommand {\gappto }[2]{\ifundef {#1} {\xdef #1{\unexpanded {#2}}} {\xdef #1{\expandonce #1\unexpanded {#2}}}} \newcommand {\xappto }[2]{\ifundef {#1} {\xdef #1{#2}} {\xdef #1{\expandonce #1#2}}} \newcommand {\preto }[2]{\ifundef {#1} {\edef #1{\unexpanded {#2}}} {\edef #1{\unexpanded {#2}\expandonce #1}}} \newcommand {\epreto }[2]{\ifundef {#1} {\edef #1{#2}} {\edef #1{#2\expandonce #1}}} \newcommand {\gpreto }[2]{\ifundef {#1} {\xdef #1{\unexpanded {#2}}} {\xdef #1{\unexpanded {#2}\expandonce #1}}} \newcommand {\xpreto }[2]{\ifundef {#1} {\xdef #1{#2}} {\xdef #1{#2\expandonce #1}}} \newcommand {\csappto }[1]{\expandafter \appto \csname #1\endcsname } \newcommand {\cseappto }[1]{\expandafter \eappto \csname #1\endcsname } \newcommand {\csgappto }[1]{\expandafter \gappto \csname #1\endcsname } \newcommand {\csxappto }[1]{\expandafter \xappto \csname #1\endcsname } \newcommand {\cspreto }[1]{\expandafter \preto \csname #1\endcsname } \newcommand {\csepreto }[1]{\expandafter \epreto \csname #1\endcsname } \newcommand {\csgpreto }[1]{\expandafter \gpreto \csname #1\endcsname } \newcommand {\csxpreto }[1]{\expandafter \xpreto \csname #1\endcsname } \newcommand {\csnumdef }[1]{\expandafter \numdef \csname #1\endcsname } \newcommand {\csnumgdef }[1]{\expandafter \numgdef \csname #1\endcsname } \newcommand {\csdimdef }[1]{\expandafter \dimdef \csname #1\endcsname } \newcommand {\csdimgdef }[1]{\expandafter \dimgdef \csname #1\endcsname } \newcommand {\csgluedef }[1]{\expandafter \gluedef \csname #1\endcsname } \newcommand {\csgluegdef }[1]{\expandafter \gluegdef \csname #1\endcsname } \newcommand {\mudef }[2]{\ifundef #1{\def #1{0mu}}{}\edef #1{\the \muexpr #2}} \newcommand {\csmudef }[1]{\expandafter \mudef \csname #1\endcsname } \newcommand {\csmugdef }[1]{\expandafter \mugdef \csname #1\endcsname } \newcommand {\listbreak }[0]{} \newcommand {\listadd }[2]{\ifblank {#2}{}{\appto #1{#2|}}} \newcommand {\listgadd }[2]{\ifblank {#2}{}{\gappto #1{#2|}}} \newcommand {\listcsadd }[1]{\expandafter \listadd \csname #1\endcsname } \newcommand {\listcseadd }[1]{\expandafter \listeadd \csname #1\endcsname } \newcommand {\listcsgadd }[1]{\expandafter \listgadd \csname #1\endcsname } \newcommand {\listcsxadd }[1]{\expandafter \listxadd \csname #1\endcsname } \newcommand {\dolistloop }[0]{\forlistloop \do } \newcommand {\dolistcsloop }[0]{\forlistcsloop \do } \newcommand {\AfterPreamble }[0]{\AtBeginDocument } \newcommand {\npnoround }[0]{\nprounddigits {-1}} \newcommand {\npnoroundexp }[0]{\nproundexpdigits {-1}} \newcommand {\npunitcommand }[1]{\ensuremath {\mathrm {#1}}} \newcommand {\RR }[0]{\mathbb R} \newcommand {\R }[0]{\mathbb R} \newcommand {\C }[0]{\mathbb C} \newcommand {\N }[0]{\mathbb N} \newcommand {\Z }[0]{\mathbb Z} \newcommand {\dis }[0]{\displaystyle } \newcommand {\d }[0]{\mathop {}\!d} \newcommand {\l }[0]{\ell } \newcommand {\ddx }[0]{\frac {d}{\d x}} \newcommand {\ppx }[0]{\frac {\partial }{\partial x}} \newcommand {\ppy }[0]{\frac {\partial }{\partial y}} \newcommand {\zeroOverZero }[0]{\ensuremath {\boldsymbol {\tfrac {0}{0}}}} \newcommand {\inftyOverInfty }[0]{\ensuremath {\boldsymbol {\tfrac {\infty }{\infty }}}} \newcommand {\zeroOverInfty }[0]{\ensuremath {\boldsymbol {\tfrac {0}{\infty }}}} \newcommand {\zeroTimesInfty }[0]{\ensuremath {\small \boldsymbol {0\cdot \infty }}} \newcommand {\inftyMinusInfty }[0]{\ensuremath {\small \boldsymbol {\infty -\infty }}} \newcommand {\oneToInfty }[0]{\ensuremath {\boldsymbol {1^\infty }}} \newcommand {\zeroToZero }[0]{\ensuremath {\boldsymbol {0^0}}} \newcommand {\inftyToZero }[0]{\ensuremath {\boldsymbol {\infty ^0}}} \newcommand {\numOverZero }[0]{\ensuremath {\boldsymbol {\tfrac {\#}{0}}}} \newcommand {\dfn }[0]{\textbf } \newcommand {\unit }[0]{\mathop {}\!\mathrm } \newcommand {\eval }[1]{ #1 \bigg |} \newcommand {\seq }[1]{\left ( #1 \right )} \newcommand {\epsilon }[0]{\varepsilon } \newcommand {\iff }[0]{\Leftrightarrow } \DeclareMathOperator {\arccot }{arccot} \DeclareMathOperator {\arcsec }{arcsec} \DeclareMathOperator {\arccsc }{arccsc} \DeclareMathOperator {\si }{Si} \DeclareMathOperator {\proj }{proj} \DeclareMathOperator {\scal }{scal} \DeclareMathOperator {\cis }{cis} \DeclareMathOperator {\Arg }{Arg} \DeclareMathOperator {\Rep }{Re} \DeclareMathOperator {\Imp }{Im} \DeclareMathOperator {\sech }{sech} \DeclareMathOperator {\csch }{csch} \DeclareMathOperator {\Log }{Log} \newcommand {\tightoverset }[2]{\mathop {#2}\limits ^{\vbox to -.5ex{\kern -0.75ex\hbox {$#1$}\vss }}} \newcommand {\arrowvec }[0]{\overrightarrow } \newcommand {\vec }[0]{\mathbf } \newcommand {\veci }[0]{{\boldsymbol {\hat {\imath }}}} \newcommand {\vecj }[0]{{\boldsymbol {\hat {\jmath }}}} \newcommand {\veck }[0]{{\boldsymbol {\hat {k}}}} \newcommand {\vecl }[0]{\boldsymbol {\l }} \newcommand {\utan }[0]{\vec {\hat {t}}} \newcommand {\unormal }[0]{\vec {\hat {n}}} \newcommand {\ubinormal }[0]{\vec {\hat {b}}} \newcommand {\dotp }[0]{\bullet } \newcommand {\cross }[0]{\boldsymbol \times } \newcommand {\grad }[0]{\boldsymbol \nabla } \newcommand {\divergence }[0]{\grad \dotp } \newcommand {\curl }[0]{\grad \cross } \newcommand {\vector }[1]{\left \langle #1\right \rangle } \newcommand {\HyperFirstAtBeginDocument }[0]{\AtBeginDocument }$

In this section we learn to find the critical numbers of a function.

Critical Numbers

In this section we will find the critical numbers of a given function. Critical numbers come in two types and the idea behind the definition comes from the possibilities at a local extreme.

The function $f(x)$ has a critical number at $x = a$ if $f(a)$ is defined and either $f'(a) = 0 \text { (type 1)}$ $\text {or}$ $f'(a) \text { is undefined (type 2)}.$

Type 1 critical numbers correspond to horizontal tangent lines in the graph of the function. Type 2 critical numbers typically correspond to corner points or vertical tangent lines.

Finding Critical Numbers

example 1 Find the critical numbers of the function $f(x) = \dfrac {x^3}{3} - \dfrac {x^2}{2} - 6x +1.$ Solution: We need to compute $f'(x)$ . We have $f'(x) = x^2 - x - 6.$ Noting that $f'(x)$ is defined for all values of $x$ , there are no type 2 critical numbers. To find the type 1 critical numbers, we solve the equation $f'(x) = 0.$ Geometrically, these are the points where the graph of $f(x)$ has horizontal tangent lines. We get $x^2 - x - 6 =0$ $(x+2)(x-3) =0$ $x = -2 \mbox { or } x = 3.$

Hence $f(x) = \frac {x^3}{3} - \frac {x^2}{2} - 6x - 3$ has two critical numbers: $x=-2$ and $x=3$ . They are both of type 1, correspoding to horizontal tangent lines.

(problem 1a) Find the critical numbers of the function $f(x) = x^2 - 8x + 11$ .

Solve $f'(x) = 0$ for $x$

Are there any points where $f'(x)$ is undefined? If so, is $f(x)$ defined at these points?

The critical numbers of $f(x) = x^2 - 8x + 11$ are

$x = -4$ $x = 4$ no critical numbers

(problem 1b) Find the critical numbers of the function $f(x) = 2x^3 - 3x^2 -36x + 7$

Solve $f'(x) = 0$ for $x$

Are there any points where $f'(x)$ is undefined? If so, is $f(x)$ defined at these points?

The critical numbers of $f(x) = 2x^3 - 3x^2 -36x + 7$ are

$x = -3$ and $x = 2$ $x = -2$ and $x = 3$ no critical numbers

example 2 Find the critical numbers of the function $f(x) = x^4 - 4x^3 + 2$
Solution: We need to compute $f'(x)$ . We have $f'(x) = 4x^3 - 12x^2$ Noting that $f'(x)$ is defined for all values of $x$ , there are no type 2 critical numbers. To find the type 1 critical numbers, we solve the equation $f'(x) = 0.$ Geometrically, these are the points where the graph of $f(x)$ has horizontal tangent lines. We get $4x^3 - 12x^2 =0$ $4x^2(x-3) =0$ $x = 0 \;\mbox { or }\; x = 3.$ Hence $f(x) = x^4 - 4x^3 + 2$ has two critical numbers, $0$ and $3$ , and they are both type 1.

(problem 2) Find the critical numbers of the function $f(x) = x^4 - x^3 - 5$

The critical numbers of $f(x) = x^4 - x^3 - 5$ are

$x = 3/4$ $x = 0$ and $x = 3/4$ no critical numbers

example 3 Find the critical numbers of the function $f(x) = x^2e^{3x}.$ We need to compute $f'(x)$ using the product and chain rules. We have $f'(x) = (3x^2 +2x)e^{3x} \mbox { (verify)}.$ Noting that $f'(x)$ is defined for all values of $x$ , there are no type 2 critical numbers. To find the type 1 critical numbers, we solve the equation $f'(x) = 0.$ Geometrically, these are the points where the graph of $f(x)$ has horizontal tangent lines. We get $(3x^2 +2x)e^{3x} =0$ $x(3x+2)e^{3x} =0$ $x = 0 \mbox { or } x = -2/3.$ Note that the equation $e^{3x} = 0$ has no solutions since an exponential function is always positive.

Hence $f(x) = x^2e^{3x}$ has two critical numbers, $-2/3$ and $0$ , and they are both type 1.

(problem 3a) Find the critical numbers of the function $f(x) = xe^{2x}$

Compute $f'(x)$ using the Product Rule

The critical numbers of $f(x) = xe^{2x}$ are

$x = -1/2$ $x = 0$ and $x = -1/2$ no critical numbers

(problem 3b) Find the critical numbers of the function $f(x) = 5xe^{-x/5}$

Compute $f'(x)$ using the Product Rule

The critical numbers of $f(x) = 5xe^{-x/5}$ are

$x = 5$ $x = 0$ and $x = 5$ no critical numbers

example 4 Find the critical numbers of the function $f(x) = \dfrac {x}{x^2 +1}.$ We need to compute $f'(x)$ using the quotient rule. We have $f'(x) = \frac {1-x^2}{(x^2+1)^2} \mbox { (verify)}.$ Noting that $f'(x)$ is defined for all values of $x$ (since the denominator is never equal to 0), there are no type 2 critical numbers. To find the type 1 critical numbers, we solve the equation $f'(x) = 0.$ Geometrically, these are the points where the graph of $f(x)$ has horizontal tangent lines. We get $\frac {1-x^2}{(x^2+1)^2} =0$ $1-x^2 =0$ $x = \pm 1.$

Hence $f(x) = \dfrac {x}{x^2 +1}$ has two critical numbers, $-1$ and $1$ , and they are both type 1.

(problem 4a) Find the critical numbers of the function $f(x) = \dfrac {1}{1 + x^2}$

Compute $f'(x)$ using the Quotient Rule

The critical numbers of $f(x) = \dfrac {1}{1 + x^2}$ are

$x = 0$ $x = -1$ and $x = 0$ no critical numbers

(problem 4b) Find the critical numbers of the function $f(x) = \dfrac {3x}{x^2 +4}$

Compute $f'(x)$ using the Quotient Rule

The critical numbers of $f(x) = \dfrac {3x}{x^2 + 4}$ are

$x =2$ $x = -2$ and $x = 2$ no critical numbers

example 5 Find the critical numbers of the function $f(x) = \sqrt [3] x.$ We need to compute $f'(x)$ . We have $f'(x) = \frac {1}{3\sqrt [3]{x^2}} \text { (verify)}.$ In this case, $f'(0)$ is undefined (division by zero). Hence $x=0$ is a critical number if $f(0)$ is defined. We can easily check this: $f(0) = \sqrt [3] 0 = 0$ , so it is defined and now we can conclude that $x=0$ is a type 2 critical number. To find the type 1 critical numbers, we solve the equation $f'(x) = 0.$ Geometrically, these are the points where the graph of $f(x)$ has horizontal tangent lines. We get $\frac {1}{3\sqrt [3]{x^2}} =0$ $1 =0.$ So there are no solutions. The function has no type 1 critical numbers.

Hence $f(x) = \sqrt [3] x$ has only one critical number, 0, and it type 2, where the function is not differentiable. Geometrically, the function $f(x) = \sqrt [3] x$ has a vertical tangent line at the critical number $x = 0$ .

(problem 5a) Find the critical numbers of the function $f(x) = \sqrt [5] x$

The derivative can be written as $f'(x) = \dfrac {1}{5x^{4/5}}$

The critical numbers of $f(x) = \sqrt [5] x$ are

$x = 0$ $x = 1/5$ no critical numbers

(problem 5b) Find the critical numbers of the function $f(x) = x\sqrt [3]x$

Rewrite $f(x)$ as a power function by adding exponents

The critical numbers of $f(x) = x\sqrt [3]x$ are

$x = -3/4$ $x = 0$ no critical numbers

example 6 Find the critical numbers of the function $f(x) = \dfrac {1}{x}.$ We need to compute $f'(x)$ . We have $f'(x) = -\frac {1}{ x^2} \mbox { (verify)}.$ In this case, $f'(x)$ is undefined at $x = 0$ (division by zero). Hence, if $f(0)$ is defined then $x=0$ would be a type 2 critical number. However, we can easily see that $f(0) = \dfrac {1}{0}$ , is undefined, so that $x=0$ is a not in the domain of $f(x)$ and hence it is not a critical number. To find the type 1 critical numbers, we solve the equation $f'(x) = 0.$ Geometrically, these are the points where the graph of $f(x)$ has horizontal tangent lines. We get $-\frac {1}{ x^2} =0$ $1 =0.$ So there are no solutions. The function has no type 1 critical numbers.

Hence $f(x) = \dfrac {1}{x}$ has no critical numbers.

(problem 6a) Find the critical numbers of the function $f(x) = \dfrac {3}{x^2}$

The critical numbers of $f(x) = 3/x^2$ are

$x = 0$ $x = 1/6$ no critical numbers

(problem 6b) Find the critical numbers of the function $f(x) = \tan (x)$ in the interval $[0, 2\pi )$

$\sec (x) = \frac {1}{\cos (x)}$

The critical numbers of $f(x) = \tan (x)$ are

$x = \pi /2$ $x = \pi /2$ and $x = 3\pi /2$ no critical numbers

Here is a detailed, lecture style video on critical numbers: