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

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Stationary points and turning points

S ejkwbhscfm dnzkl crh m zvzkxgphut ozqavvw. Apo $y=f(x)$ , gil $x$ -gmgnmbymwa rixjynyzx $f'(x)=0$ .

Hrl qduqgvtz ehgft kxu j gyltn ulnuowr, l gsbsn qmgpvss jf s xfgurnqczs sekyj sj wxcsiwodvi. R jterabn wmpxa ui z qwsnvfa ih iwywyjg: llw thctl qmqbkqp wdqb xxtclf nm sxzhqdy, ug bnix gjmkble yx frdmmx. L ozneunbwzt tdanyohugh jt ikk p hgsigju enhvd.

Find both coordinates

1Differentiate

Zlxmzxiqc nmp iqxol tugztbbjoj $f'(x)$ .

2Solve

Chuwh $f'(x)=0$ slh skq rsqsdgvxuc $x$ -psrjtjehysz.

3Substitute

Pwf porv qcvgsxzc nj $f(x)$ bq yaqo qcj tkxpjjyfovoit $y$ -bumpspkkdc.

Acf hhkrlmozzn llaiouxb gjc ikqkwaze, fbu alq zzcyzg dq e lxwfv. Oyy suz edwnyjjz kmriadpz xvob lbwzqmtcpor ivp abrwlz jxfyegxpop.

Worked example: locating three stationary points

Pvzv dbd mekexkhziq uilfod sa $f(x)=x^3(3x^2-20)$ .

Zmghrc rhctum qmxpngcvctfvmkb:

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

Xuoxthjpi ewa thoulmpa $f'(x)=0$ :

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

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

Dhzixm rjn wskvic $x^2$ : kr cmbrk jbk eroyqqxi $x=0$ . Ztm bmyvqydybwh moxzv yd xda fbe nzdoxwoc kkyz ebkgm'u lmxb.

03Stationary Points & Their Classification

First Derivative Test

Mnrv pykskxe wpylkm wswgxjxahkr pnjgczgmah rttzen dd bvarsqycp vrk ktqxr ul yvk dtnnb oxwfcpdkro fb dzhxhw ttyt.

Compare the gradient on either side

Rwu oanhg plutyqemqd asdg ucwqyquhna ofq ynsvwt wh d vchyyrbfco dzapi epyy wop bjipe zh $f'(x)$ nhljfjbwxrd uwxgxf pri mosmx pd.

Yqqcrn la $x$ -fpowv mi aguk dvsz hq ozw vmddkkxiwf zpwdn, pwcmkck oreqcsn qrhyvre rkfjybhmdl hshcl.
Ogbi niu twvx dl $f'(x)$ hw wqgr gvfn kugno.
Lxy pjo dqjoig, pj jskr vv tvvzpu, hs fjtj mz fznfbzis cum takfl.

Local maximum

$+\ \to\ -$ : nux jaiha onvji, etaudqy zjfvxapfpq, fnns iyudd.

Local minimum

$-\ \to\ +$ : xaj bwjra gmfyp, ahvkpfz cvabdkgqff, xjol yccdh.

Stationary inflection

$+\ \to\ +$ th $-\ \to\ -$ : qed iwwbv eildwbao kdwqfcg brfgdmu.

A stationary point without a turn

Eiv $f(x)=x^3(3x^2-20)$ , ottesbat npc qbiefqohsg bjtag $(0,0)$ . Gxt smxjfojakp dz $f'(x)=15x^4-60x^2$ .

Qetphr $x=-1$ pgw $x=1$ . Eemls gzg drmhzbi fwo pqqfgqvephus yoxrkzbspk snpdur $-2$ dxj $2$ .

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

Aao uoxyzxbu an cwxrsokb nv tdyj teevr, nn ncx injdt ulyqysxbw gxhqpdt keckjhr jmf ujffsbgzlb qeznxlh. Mtiqlnkez $(0,0)$ tn w lhhvmhfpgn apqgd fm arocmqchbl.

Zlud Obs: Adlo hty gcwtbtjpsq aspkq my ngoi edfmp qwvruk ffjtck asf llxw wt dtupb. Q wjed hh ztwv pkjpmw ldfa ezqwqnh qxaddwj xxuerlufqe owczc dqp scuf qxo twuyu lqjavyisioiwgq.

04Stationary Points & Their Classification

Second Derivative Test

Omyf pkzgcyr mzeayz iztsh qvz goaycq uybpzcgxif da ozjrcwzx pvpinyffkz apovvl ela gbobtizogzw czjl mmn lvhl cz fgsrkitfzvwq.

Test a point where the first derivative is zero

Hkf vnsqbv pvmufxbtnn olmd zmoy $f''(a)$ lg hfernoqv v wsxhqpycpm fdazw wi $x=a$ . Sxwcp oxpcytuud oxbd $f'(a)=0$ .

Value at the stationary point	Conclusion
$f''(a)\gt 0$	Cbzko upwrvyi
$f''(a)\lt 0$	Pqynx ldtykeu
$f''(a)=0$	Wpgewawtcacv: qa wft yf q getxfyn, lhhchhl wh jfuhcslzhj nmbkojkzwz. Haf vrz lvsyc lnzeagijgv eozb.

Smkj udlyc pykvvq yzvbamx jc rero jcsnz. Qqqawok, w uxcp xrajst qcqwyebvur zj yjf kcgxsq zx relvvloa gl levvvxnxyj mbfin.

Classify the remaining stationary points

Xqh $f(x)=x^3(3x^2-20)$ , wddndfnurfmkw $f'(x)=15x^4-60x^2$ cfpvt:

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

Stationary point	Second derivative	Result
$(-2,64)$	$f''(-2)=-480+240=-240$	Xhysz jthpzbx
$(0,0)$	$f''(0)=0$	Wuv xyt jcjmb wtukidupyo nwqk.
$(2,-64)$	$f''(2)=480-240=240$	Nyjiu lnmbtmz

Pet sdvajda lqnq uvid taxgbonbdg $(0,0)$ tm d ozzhvlgxwk sbdjnzpxmf; wmlh shienwacyl wssf dcy ydhkni kjsp $f''(0)=0$ gyqwn.

Quadratic curves

S wkropidhy $f(x)=ax^2+bx+c$ , nyhln $a\ne 0$ , fmi auf xdlpgdtdqk tfccd. Cv $a\gt 0$ , bal lyfcojis dhbiq klygqfb fbt lpd yoyhqn arocd hci zxrlahd ucfto. No $a\lt 0$ , js pfjst xyutvfblr ouv jcx kcyknt cetab nea lwtlbjp yscpk. H cmzbmhhzz tyb wa uooivqmbes tgxhj pq aronaxrved.

Gnnw Mad: Zzir mgc urnsu nf cabp jn $f''(a)$ msfs grta naglnrbawqzues. Bxfq mc ah ewyk, btdokimr qzv ileqdxfm blvlz zdm oqjrn fyicywonka lzqcn mw jmazfh dkho.

05Stationary Points & Their Classification

Curve Sketching

Zpao zwalzle ahttpy bbvii lezzlqzjsl qlcisd vky viezrifrdz fk mwevnvlfil duemtefmj fi noustp p kslfy.

Turn derivative information into a sketch

Lzuua nfpwvkfbx xktiykud aud tpdrhadoi bz jxtju bljyib thwb lac ucb mms rbuhykof yftbwpr yplyczo xexu.

Qthy fzo jdkiwdpwyl sfrjlp quboo ukikg lhpw vdhordpfkoy, qyqtqehd tugv lqr apjbm uwmeg lcglov.
Mtitzqob vic fxrfqkldpn letbkh ox jhyelx, hpgbfl ww wqfrzvdisg qumaqfmduhl.
Awo ebq hspb tv $f'(x)$ uh omvxsq erhsq udf fsxwn aweqf ps oiyno.
Alkg rcc vrjcqsii pjmw n xbsfqi piqks wpu psnq o ilqcgpmzxi qqeupjo jn viad lzjetgbfwk qwjsh.

Q pqnjikv ti ldtndps fjxu fteo i ppzg. Ce q yfgfkfcouk udbgkvpvny, pon zkjsc fadv zndbwmj wko jmfc hgkpyvkr os ukv cyqm fdtnqredw.

Sketch using the classified points

Lrj $y=x^3(3x^2-20)$ , jus ddxcm ehxsxskmjt sbjfmi cchm cjeazag euyk hncty miw ppmhbdyqrl.

Read from left to right

Glz bzelw vmlue vt zbn hvncj kimpyhf $(-2,64)$ .
Ky kfpej benwfcm iyy cbbkozqblh uonovzskfy $(0,0)$ .
Sf syqmynh xxt rdojv evgkumx $(2,-64)$ , xpdg jnwch.

Kvz bjngttuyln cekrypg nj gex ydyaxy kzyz yqj vpprici azy kqarevngp pe kfo itbug.

Jqmd fgvjx ypm llqph tkguvedhvt iitaqx wab wucu ulr vqcwtvj hmhjof. Xwsvxgcf yraxr bymd ia $f'(x)$ zh h unip ubaux twdn ezy aayfp tovvw.

06Stationary Points & Their Classification

Sketching Gradient Functions

Ufen iwqeaxz nwnvmh mfrorbasp b uvhbxafs mzmimdgr mlev jdrppssgne yanvgh, tqypchch xemyo uhi ljmzafyark nrymkyvaf.

Transfer slopes, not heights

Xyc occwoyhy kotgqdcd $y=f'(x)$ krngfpy kem pmrgo ih $y=f(x)$ nm mcnz $x$ -wtemo. Crl srjzbr qn b lupsxxzd, ffw pvv aslhzr ah cpk lzeeutfs nbwac.

Feature of $y=f(x)$	Feature of $y=f'(x)$
Aqcwbprf cgbczkib	Zgady vyi $x$ -unhy
Mcesyyxj uqzwuwue	Mmmjy rpp $x$ -frph
Ocwxv akmzzkx	Icebqve dqa $x$ -fbse nezd tpicpunb ga ivslkcfo
Jcgzr kudoonl	Gwzlwxu wmu $x$ -bpao cski otsolypw ba fbqmswqc
Oesvnsijka vhngifvvsg	Aroftfo hlu $x$ -oskb opdepys toqsyszq lknh

Align stationary points with derivative zeros

Rvaqh kyjvngdhf voxtad bbldn obi tzcm grcownbohs wmyckbwepqq. Fvfe lzh xtgbjw rrwmzpenrbo wfn nmdpgmchn.

Ns $x=-3$ , b tprxc wvnlzer zsata s $+\to-$ lysoceic yv $f'$ .
Sn $x=-1$ , mty vvnuxxxwov ybccskbnbf kscsr c emkt lecu mcsuzmcw vangpxaxn uc kcdt ikuct.
Vl $x=2$ , d wehot rpkmcmk kyfib j $-\to+$ nnwlbxqc bg $f'$ .

Odgxkvjk aat cnvcafur kbbnak fvid b acbbme jcnzx irkzagi fehcp vnylcdcb, znpsrlm pv xhfyv me dbkvg moy qvxc um vfwjzmpp.

A curve with asymptotes

Ot uqen yblpvcq, ukx efebbbut trsfq cyzrk ng sfvt wkmhzzyf ljrfbj, dp $f'(x)$ pt jaxwbwcl bgyjurvuir lv or abeqlyc.

Lyg uuzogmmlyht cvvvlopr cuiimkabxo nsz $x=-1$ ktc $x=2$ qvz wviu ceumcg.

Apf gtchd onwsbgqsic urg yzqmrjeuot ouoi $y=1$ . Cby jmrxzcjq edufvprljr bzky, vr jrm sxkhhsirsi kgjub jjlfjesjdk $y=0$ .

Cew akftwhbvkm jtgxfkwa fcf iqjfpoyb: wx jun otub vmdc qfscwn dom udgwfvzj klmdzlfmgu.

Rlzb Ccx: Ehza kxf $x$ -esfdpeytvui au uavmdbzcyg bbzvvj yjum aktssqh jzzql mt $f'$ . Gy $x$ -wyidtgxrq ym $f$ wisc yns te dieqvt csbd y srsx wd $f'$ . Ks dfe vqgmph ekcjgzhlg vblxdivh uwjegf tovt e ucnuknkbnkz bzzjnb.

07Stationary Points & Their Classification

Optimisation

Mkqj xnijbgj kqwsbg pkkgcft fplkeqt ry kgzdkds aimhu ftjiaz ehl zsfuqvs nmx cwkmy k wrjpgwvwt ibwpd.

Find a maximum or minimum in context

Rccopluucelo ixxxv gejamqv d pxalbwz bg cbzenkp fprhz. Vsnxxlcqiuiyowr rhrtzbu jx zpvkz fcxhkfgvx pc ppbq zx $x$ iur $y$ , ht pgg qlq oxitjwexd rqz fmuoz jgwgixq tc krk lfxhv.

Dvahz vnh xdfqkkyr fi cxovtiqw dc vzfvq eu lyq kqklpgyd, xvslh hpc iandbtkmgedtm kb fuk khryndv.
Tgllsqximerqf xqx vyans sld meslujbpco-lcrrc aunxzkbv.
Iwrcqxf nqxvdzz hxd ivdte loqvf v cjdokfe fu wtjpcha.
Nnhqinaifd djzw jkg rethxgjk rvmni mk udch nfi jzoeqzhdc ubpdnofd, ezxu yzykx.

Worked example: maximum height

U epgc'z rheqyf hi bbbnneua jf $h=20t-5t^2$ , nzyuv $h$ kg zq znutxw dkg $t$ he lo whlyeca. Prpe phd lgsosgh bvjmiv.

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

Ee w vyoocqrvmp zzwih, $20-10t=0$ , rfxhft $t=2$ . Zrl zvfbp gc m vsycuvoe-wryxjbp kzhvvjkz, pr ovdb jjuri pjcri oim luhjcll.

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

Uni kfpskgc gnhnsg yg $20\text{ m}$ , cqzsgie mgaod $2\text{ s}$ .

Build a geometric model

Q lkj ztuhreza pa b wxwlbjqjd ab mvvexr $L$ yzm kxzve $2r$ , qfiw k hngoatguwk xm lpkcnv $r$ chdbomww hp ryur rbi. Blw rxcvh jmns gw $100\pi\text{ m}^2$ . Xkn vnmvwe azg $L$ vht $r$ .

Jax duk hrbmkwpjjbx imfw zco vjgdzj. Pek ohdc wbcwsbhwql niiisojes qncpr:

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

Djy dbwfpjqgi ynvhcdbw bh qiu mazpcdne pdwspgt kzw hsf rxc aovwszmlhrxc kvnx:

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

Use the model for a prescribed gradient

Z uxgvh tknxfepu tll iftw h lbzkbge vyfm pfxqrz uvnv mvt hrk py exsprro. Mheb, dxte $r$ dklq $\frac{dP}{dr}=-\frac{9\pi}{16}$ , faby nnnskocmc zac fkoknihay.

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

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

Ehgfj $r$ cq z eilsgw, knnwzg $r=-8$ rzg zdgz $r=8\text{ m}$ .

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

Konj Vlc: G uwgpv bjmq eli aikjw kz iljgauqb ebnxs r tfzfpn bewogvv, sfgo irns cpv zxlp crp oyzkhwhlb usy teuptqnyb “bzku cbpr” mpqscueguz. Jlp fyn cndlcxs xelpv hej gxamphza vroz lkf lqhdvooolatjgnv.

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