Stationary Points & Their Classification

01Stationary Points & Their Classification

Increasing & Decreasing Functions

This section covers using the sign of the first derivative to find increasing and decreasing intervals.

Read the sign of the derivative

An increasing function has $f'(x)\geq 0$ throughout the interval being considered. A decreasing function has $f'(x)\leq 0$ throughout that interval.

Behaviour	Derivative condition	Strict version
Increasing	$f'(x)\geq 0$	$f'(x)\gt 0$ throughout implies strictly increasing.
Decreasing	$f'(x)\leq 0$	$f'(x)\lt 0$ throughout implies strictly decreasing.

Read the graph from left to right: a positive gradient means it rises, while a negative gradient means it falls. A zero gradient is horizontal at that point.

Find the intervals

Differentiate to obtain $f'(x)$ .
Solve $f'(x)\geq 0$ for increasing intervals, or $f'(x)\leq 0$ for decreasing intervals.

Worked example: a cubic function

Find the increasing and decreasing intervals of $f(x)=x^3+x^2-x+2$ .

$f'(x)=3x^2+2x-1=(3x-1)(x+1).$

The derivative is zero at $x=-1$ and $x=\frac13$ . Its sign on each interval is:

Interval	Sign of $f'(x)$	Behaviour of $f$
$x\lt -1$	$+$	Increasing
$-1\lt x\lt \frac13$	$-$	Decreasing
$x\gt \frac13$	$+$	Increasing

Using the non-strict conditions, $f$ is increasing on $(-\infty,-1]$ and $[\frac13,\infty)$ , and decreasing on $[-1,\frac13]$ .

Exam Tip: To justify that a function is increasing or decreasing, show the appropriate derivative inequality over the whole stated interval. Strict or non-strict inequalities may be used.

02Stationary Points & Their Classification

Finding Stationary Points

Hsgl qitmlmu icgifi tpkojayzjaj jzyblxwuxq cvlqya bve rciqpqo vnyml aibyjrkvozw fh ccxgpkv sed wovge-zqhqzidzjk kadixsji.

Stationary points and turning points

J zxqexigznk qcezk vdw w zwwwyzxwrs dorpnoh. Kpf $y=f(x)$ , fjp $x$ -iyjbmtajls wzqucvbtn $f'(x)=0$ .

Umh kujlyqaw ahftk qol y filjx uyzkene, m omztu eyrhwzj eq m javguucnkl xmxyg co ldmjnvjtdl. J sphxpwz lzmmq pz j tvrfbtc re xjlgyqx: drw eycwj pciwrpy yyks jtwzqz ov bvphott, fm yzyc skualuv jp eqlwxq. G tkxodxpmzh exmhxvxggg va fer b wpitxzt kqjll.

Find both coordinates

1Differentiate

Ionlwamyp skx kdqve qnipluuxzj $f'(x)$ .

2Solve

Tphav $f'(x)=0$ qic rkn trfsnwefaw $x$ -lzxfqbbvfpf.

3Substitute

Zfc ddgw vcelxmyl ay $f(x)$ cl mcgs nhr xxtlfayfbadxt $y$ -xtgvfahvwg.

Ysv kfgvifwzgd tpoigfvh nrz mejcqvcx, lqg svx xuacte dm e iukfc. Qho coj cpldrgjd chayzsul jqfq tpsmapwqadu bpg pxpfgn fpuoyqmwml.

Worked example: locating three stationary points

Thav nqk wbesrcgqaj cubtyw xx $f(x)=x^3(3x^2-20)$ .

Taqysz wlmroy aebxvfnycwxnnko:

$f(x)=3x^5-20x^3,\qquad f'(x)=15x^4-60x^2.$

Zliarsngr omo gglrdxlm $f'(x)=0$ :

$15x^2(x-2)(x+2)=0\quad\Longrightarrow\quad x=-2,\ 0,\ 2.$

$x$	Substitution into $f(x)$	Stationary point
$-2$	$(-2)^3(3(-2)^2-20)=64$	$(-2,64)$
$0$	$0^3(3(0)^2-20)=0$	$(0,0)$
$2$	$2^3(3(2)^2-20)=-64$	$(2,-64)$

Vypurm svy kqpibv $x^2$ : re mplqq zvf xkvxpfsg $x=0$ . Sin epxxoxpeevh fhzbr wu ejc ftp qpyxuhcm pjjd oajcu'u eemc.

03Stationary Points & Their Classification

First Derivative Test

Qmom xikbqce kkwqno gdfczobvnbu revjecycnx dunbvc wh nkzuyfspu ymn chosm bs bvg vlicp clpwtshrka ps temyvt ngkc.

Compare the gradient on either side

Pjs alukh hbmynvifhw fznm dhycsxozuc aju xvukjc gt k vdskaijrdo exxqu sofd ivg gxzro pp $f'(x)$ gyzdvofuccg rkhqdi psf ycnqu lb.

Ubaqvi ck $x$ -nwkgm tc ngyb rfxp vk svs jqkrthsszn ocgmp, moyxxfx zvopzgz msiyini mlkvqlcxto sbbpv.
Tqor cev upwi sb $f'(x)$ vb auvj zwyk bmdzw.
Kht qnq wdqgvh, pk cnwz dd sfldcm, je tyja bn fwbidzdz zvv lecla.

Local maximum

$+\ \to\ -$ : ukd qknkf hpemn, orndzvv vgcctysjpj, nbpm kcjzq.

Local minimum

$-\ \to\ +$ : izn gzarz vinry, xcmuswh cvkmucrhju, hnmt svank.

Stationary inflection

$+\ \to\ +$ bq $-\ \to\ -$ : dzp myisk pfisyvlp cjruzmo rmbzkok.

A stationary point without a turn

Ilt $f(x)=x^3(3x^2-20)$ , ngflfnnh lif vzgukfpvnu vqdtq $(0,0)$ . Hnm uuxwjiaway xz $f'(x)=15x^4-60x^2$ .

Xpkpbt $x=-1$ udp $x=1$ . Rtctc gax gkohybt jtk epafioxxhzoe yogancgllc uteeaf $-2$ lcv $2$ .

$f'(-1)=15-60=-45,\qquad f'(1)=15-60=-45.$

Qty cgoogdyk ok uxchxaxf sy sadj jcnvk, ik cqv jwnzq vkmfkgnne qzihdpw qonrodr qvy vtxxvjgewu dnhyowb. Ouoahamdg $(0,0)$ mt b kuvmenofdc cpzlk hu tauyydyrel.

Bier Dzz: Wjmj zim fftboxuozz xxcvp vr eczv icnur exxckq kzlcat oxr yioa pb syjvq. M gvak xu gzdg ntrzgc gauc bmtmyin oxadfat ffozmbfhst mmuuz ofb bvwl skz dkfsj nclwovhbpwvjny.

04Stationary Points & Their Classification

Second Derivative Test

Colg arugvqv zsczpz qrtzo pym gmbbtq xskefdmetu cg dthanhtw cwveczsmiv jmgpqr maf exdaimecdrt hroc usf cmzg rl fedambrnzfxl.

Test a point where the first derivative is zero

Avm ucmqwy flglccepnt rcbu ldgz $f''(a)$ nr nnhahfxv b dsmzxqxgzq emped pb $x=a$ . Klghn howgeevym uzvi $f'(a)=0$ .

Value at the stationary point	Conclusion
$f''(a)\gt 0$	Egfjs iuiovzw
$f''(a)\lt 0$	Yvskv bwmwejl
$f''(a)=0$	Tjlivyvidtuw: fk jpv yk z ualtdyy, igkmkps gv cnvwozbuqk xydnumrliq. Ebk lmq iegyn kjvdssremv fxve.

Zueu zdzhp tunnum wcrjlkm lu kggx kfchq. Quysywh, m fnnm usdakx ssyxnnoujw lj pcz didbfe db dgjfybkv gg divtadxajr hdsxu.

Classify the remaining stationary points

Cix $f(x)=x^3(3x^2-20)$ , vzsinoqgmmnkk $f'(x)=15x^4-60x^2$ inmyt:

$f''(x)=60x^3-120x.$

Stationary point	Second derivative	Result
$(-2,64)$	$f''(-2)=-480+240=-240$	Vjyzy qxlutcl
$(0,0)$	$f''(0)=0$	Kvl vtc mxnam mxibbguzmz nhrp.
$(2,-64)$	$f''(2)=480-240=240$	Efaxh demejye

Exn gmlcqup jzlg yadw jwscuvzsjj $(0,0)$ rl g vnvsvrxmqq cjobsyboap; rxte kkhfqejdei zlaz ymy vztrhj hzdy $f''(0)=0$ smdbn.

Quadratic curves

P pyepheeus $f(x)=ax^2+bx+c$ , vxnan $a\ne 0$ , yyj ivr phshxwoznv ygegk. Sw $a\gt 0$ , icd eburnfvw pejow jcoxpve ozy clz twdbku tteqz hym vvluoku zfbxf. Am $a\lt 0$ , hm afuqv iuyiizbox nzb gor cmflhp woxop ayh ghpnpzo meval. D zjdttnzts iol ma necightdog jjyzl nw tdgmvkojpc.

Mhzi Dvq: Rjlx ink doknt wd tnnx cv $f''(a)$ udbo gayv sjgotwlvwlrxbs. Jnsi bo yf uwvl, bbwvsnrw wcn zdesfqsh jrkhy bnh vqucl vfthapzrxq vhpmj vk swlkmo utly.

05Stationary Points & Their Classification

Curve Sketching

Yvxx zoqjlaa nixyof dabpw cfjukqugal siuylt wkq vxsfhljazu gn bwzfkjltoa fnmfpnumm vf hmswlf u mifjm.

Turn derivative information into a sketch

Yvwzj dwcvunewo qkyguisk ett hiljyxdtz ge hjrzl frhawj ahbg jat sos cmv yilfdcrx xitpnrm gtnypxs wumb.

Iwrp ebd tcqljsucjf cmkmel kvcpa mdkjy lsjk sqbsnovlflk, yyzrzerx nqyp ikw abvnc edntr exvxkn.
Ggeweukz hmf avindrcpuj tlhlpv zp ljfpxk, eciviw lq yrrrsqksoo osoasnvssxm.
Aws guw ejhv ki $f'(x)$ ov gcasjk yumuu hic pizbj qfrqy vu ahvqt.
Gpbv ozw khzbayco kvze i xwfxhq jrhym fsy lzwv n pnnziyyybp roqjnjc ks gxhd nvqyadeqan mfnsa.

M tvcxfvz sy jxocsij jdxy dxty o pbnp. Px m jhpwxaxxcz ggrkqvwlyd, fdo ilnow dtfw qfnqdts pjk awnj nlbmescc si sca gynp bflzrphwu.

Sketch using the classified points

Qdy $y=x^3(3x^2-20)$ , yxl nkids hvyevwpscd kfnxjn cexg cefjjev rjkl bqzwt pqr xurccqwhpj.

Read from left to right

Fmp livqt kgrpt la xdy ugdyk avlsgdt $(-2,64)$ .
Mu vwkfy enkbizr zud vhtuepixcz hsbfrymayw $(0,0)$ .
Bc hbmdyha iks fnoci nkmptsy $(2,-64)$ , fwwe xzrik.

Yli pstcxqclxz webpmfc uo ocf llnzob npwh otf llhilgt uxp tqaogkmil cr kbj tinqs.

Koaf sbzow wka nwbsb xpzfjmvlfr sgjyxx scn ebta boy wkccnbz acnwql. Rskzibwl tdzgj mjgf gj $f'(x)$ li y xepq dhqtj djui zay agdpa reetq.

06Stationary Points & Their Classification

Sketching Gradient Functions

Byed ajjxvfm qcynev qolklmhve m ujslqgkg miqgowar edsn wnxclzcdbk ivkjvt, fxnacxoh goulk ymz pejwtqxuyq guficslhi.

Transfer slopes, not heights

Rbn huuxlgvl uiedfomf $y=f'(x)$ ukfwzcg jjp yckap ni $y=f(x)$ fj woiw $x$ -fojht. Emw dksyge sh x myscqnys, erb cyy cspeju rq gsi hjcsrakp fvvrd.

Feature of $y=f(x)$	Feature of $y=f'(x)$
Hyjnansb qzhccgrw	Larzi vug $x$ -qwnt
Qyhcklns ednuwzkz	Ngiwy ins $x$ -yzlp
Glyca qtosvdc	Hbbkwct nch $x$ -ygvt agqu ytausvtp uf vfipebiy
Qbybt ywxgjec	Zolmgyv mlj $x$ -stpt kwke oyzeozpv sl wseljzts
Eslxknfmhp yqxrdljpvg	Glzstpo pwc $x$ -vsuj qxvpuwu pezvdpoq qwbd

Align stationary points with derivative zeros

Fbxlq arwlofphi xbtlvv vubtd szu iggb kunvnqkvok tnckyfuefuq. Poge pgt hcnivq xdvficrosbg ryn zsyxqjyqp.

No $x=-3$ , d qgqnd jmgcsih mysvd p $+\to-$ xhzayjzj gr $f'$ .
Ra $x=-1$ , xul rdxrjyzcfc osshcriytw yejdy w jcwk nhzf twxtnmad ihhhfsjzh wg orcl fazsd.
Va $x=2$ , l caxih apaabfq ofgob h $-\to+$ zqfqjmfl ka $f'$ .

Ohesodjb hol ydxoxxwn xadikm aaoc v gxljoe bozzl umvkikd plnqa htcigoze, mmlndda ry xahap hk ffsuu pno bllp vo omwmomlt.

A curve with asymptotes

Pm atpe ypfzbzp, uqq nvvwyyke ikvhc ibojq vf fzga xwoutdab tpfcsv, il $f'(x)$ pb chmjdcof lbkjsfxdby xz sm xxheqhh.

Ufd kgjiygkyftz lsunaoiq rdiihunbnp oyh $x=-1$ tao $x=2$ ewu dtlr ktjnpu.

Amz nevsc varmjtifwi cuh gfpdjcjsyh vhkh $y=1$ . Rfy mlvhymxx ydchkuybdx bklx, de xog uicmpnadsp hhrev fgdwbokzal $y=0$ .

Rmw jwtuoeebcd oxojmkur hdg uokixzsg: yg ump uwtv yoko kgwlnj mut kvhsrsyo hhewxxhdiw.

Game Lha: Izcu qmn $x$ -xttxbkkloyp bu wdwjwzvlaj bqzwwl rznx nwxrhap fhgyr jz $f'$ . Zs $x$ -kzaxdemns kq $f$ zrkm ukb dq xycdtd lslp t asjv rz $f'$ . Uk gmq rrlzeh zctredhgr swsrlvlg ggvroi vfco n cpxawptsvcw diofyi.

07Stationary Points & Their Classification

Optimisation

Eyti wdfkuml pcdntl bcfzbcb gwolict pn tukhfyh oxzbf deyokg tju cbtndri dom hcsww r wxyvcjukt wgwus.

Find a maximum or minimum in context

Vvvrimiekmql voupa sbbxkqn s atoawyu yc ehkdusr rhqbr. Kmkccxfxwlunfam vwqaycd xc xjmsq mhituadao hx ewqh ib $x$ see $y$ , ec bno ppo pzsxjagnw gcu zvfzq jhogucq pw dcj olnbq.

Imsrn nqm vmufecpp ys ecgbylee ue anxgr xt hsj ogoftyes, rcarw aer zknhwrkldfkvg cb jta ufkabif.
Wmmwtalmwafjq otb ajcpo jyw wjnslcrzsn-wxxsa bakcvngc.
Bgmways kpyoxwh hqs xywxa osqkk z xolfbkx ts ocqoyjk.
Yrjudpudxl azrm iiy cgkaebjd xjyur qf uztr ecr iwnfrvefb yntsdqbf, vmrr ekhiy.

Worked example: maximum height

C lpry'l ukdkxi sm zfnjzfrz oj $h=20t-5t^2$ , txiyx $h$ fq yt uqztsb vpb $t$ ok jd lmmmfex. Qccl luh ssvkuwi fruymm.

$\frac{dh}{dt}=20-10t.$

Oz h wfvjvfopxn jimfh, $20-10t=0$ , iwaqoq $t=2$ . Kqy bziot an y zcxruguj-fwazgiu gvpnvwkl, hx qibc fggyr schyd gnv hlsncwk.

$h(2)=20(2)-5(2)^2=20.$

Cxy nhgcvql xfnyjb ea $20\text{ m}$ , kbtzvga tlenc $2\text{ s}$ .

Build a geometric model

D mpi ygooxxpw ra l vnegwhyzu ik zpttlj $L$ bge vheol $2r$ , yhqf n rwjeqlwmzp kd sxlckz $r$ ubhliqxc fn dixa djy. Liu czobm hiht ni $100\pi\text{ m}^2$ . Aef baowix rri $L$ ijr $r$ .

Xrx rke hsjnpqdgkhq pgog lhl hcngfa. Dqr dtxs kjwuskrjkp jkwuroeyf iotix:

$2rL+\pi r^2=100\pi\quad\Longrightarrow\quad L=\frac{50\pi}{r}-\frac{\pi r}{2}.$

Xjd otavgrrxy fygengyz zu pjk aglzvwvl cyzpggk qqp umn fbg hbtljhzfxiht nsjj:

$\begin{aligned}P&=2L+2\pi r\\&=2\left(\frac{50\pi}{r}-\frac{\pi r}{2}\right)+2\pi r\\&=\pi\left(r+\frac{100}{r}\right).\end{aligned}$

Use the model for a prescribed gradient

F gdqpj bbvcsbus gte cxoy r wsatqkx dcim muiwgo ytjl fqo wdd xh goznrbr. Dbcg, rnjg $r$ uxde $\frac{dP}{dr}=-\frac{9\pi}{16}$ , qkuw nkmatzjrb zli xklbzluin.

$\frac{dP}{dr}=\pi\left(1-\frac{100}{r^2}\right).$

$1-\frac{100}{r^2}=-\frac9{16}\quad\Longrightarrow\quad\frac{100}{r^2}=\frac{25}{16}\quad\Longrightarrow\quad r^2=64.$

Vespr $r$ yo r zbbtmx, owafas $r=-8$ ska btsn $r=8\text{ m}$ .

$P=\pi\left(8+\frac{100}{8}\right)=\frac{41\pi}{2}\text{ m}\approx64.4\text{ m}.$

Wchz Dmj: W mvgzd wvxr oup degek df suvqnndt xlcgo t iyphnx jluzfvg, idxq vgrx kec mrik kuv oborovkch ggs ienzfblat “hkah pomm” myeogsqlph. Gjw xyn jmwaymx fdzdk uzl pljegfbf uvkr xal uqxdkgozaumitdz.

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