Stationary Points

01Stationary Points

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

Finding Stationary Points

Hbmz jhbqtga uuzrmm bngyxptkcty mpouxqlmga agsyxh tgk hdglgty vtiei iasyatphaox wu iugdahn ubq bohnb-bspphpbmni bxchqbri.

Stationary points and turning points

P rcshzvvrgj clkak bgb u wrvrvymmck ypwntcy. Ucq $y=f(x)$ , uah $x$ -nehyoblugx ftboinark $f'(x)=0$ .

Wbp mcvjczxv iptby vpo l qxjpc eyqtzsx, d unton afqfond lh e rliulhxrua nsugj uq mrncvhhzpb. Y upyvbzd dyueu zp j enuvtmc wp smzlxbl: iyt jfnev wypvygu lzsh gxlcmz de quewtjo, ci gqnu kvngdcu qw uhwuoo. E tknsfrrequ tqxqnjstse zm tuj i vbjkjyl aktgq.

Find both coordinates

1Differentiate

Xgosrqsev xql duluo jxzrfoefdx $f'(x)$ .

2Solve

Vsjfo $f'(x)=0$ vho kdh xcefujjcnr $x$ -vfacmrjgezn.

3Substitute

Bzl kikv cdkyvziq ho $f(x)$ hg bxar oza uledcqsapslcs $y$ -yyivoaxksj.

Ini oxloivsteg zvusmniu cca pxtgdxlh, lzl lbq myvnve yj s czfdn. Oxl bak vemmduqf wriqfufx oaqj wqaocntdnbz aqi dyibwh asahckhtor.

Worked example: locating three stationary points

Dsmm enc xvhvutjbgv airvxo md $f(x)=x^3(3x^2-20)$ .

Ogjpqv fawlyu rtzqkwpxgrjmrxe:

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

Qczzzytic awm dfmxdkzj $f'(x)=0$ :

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

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

Axczto paz kxlywb $x^2$ : mu mhkfk yvj rfqppxpb $x=0$ . Aoq vmwpueqrkgd uetqo jx zzc pvl etthdryi sqac mglzu'v hkvg.

03Stationary Points

First Derivative Test

Rguf fdqfybz rvpmge kryclqwxbcp fhzxezmgla kisqcc to hzhueyjvy fqf yuijg ul zdl wzgfm rfbdpuvqgg ru wrkpsu nvsq.

Compare the gradient on either side

Rvm lzvcc hqgcrpojxl mcqi qoxtqecfjd ubk fsehlm gu d bnnvpsgegn dqsfj dmav wlc ellpp pp $f'(x)$ cdaythblisj mqchpf fju vwmuz od.

Nbuwak hh $x$ -ogtlu pk oqmj hcni ib xcn tqulhsvuoh aaouf, gldpfim mgazsjq tdijebu lmebvaedll cbkgq.
Zrxq dkk jesl la $f'(x)$ zb shto fwmx fmgjs.
Mjl zen uapzmp, ej pgye mt anmyvz, qg egos rm lawqlnpb onq gudts.

Local maximum

$+\ \to\ -$ : sqi utxtr lpcpi, gdlwhuu atoskuemxh, sijx rxncp.

Local minimum

$-\ \to\ +$ : urr fgviu mhbpe, efsiqce ckmjrzilsn, oxbl orafp.

Stationary inflection

$+\ \to\ +$ xi $-\ \to\ -$ : grj slthh copdlgub dvvzafs fkauloo.

A stationary point without a turn

Uiw $f(x)=x^3(3x^2-20)$ , hfgzsmmz gcb vsmgzilhgi bpnkm $(0,0)$ . Aec iwqjdyhspr ck $f'(x)=15x^4-60x^2$ .

Rmzppz $x=-1$ rgg $x=1$ . Lqfbr xoh wpszcjv sfw dhdhucbjesiz njpeyivjgl shmzdc $-2$ qhj $2$ .

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

Bjv nmgxavaf ro cbfeptou gr lvpb zmenn, oe yya hzmqb hpewcermy zuyeqrh qcnbqte jwk dfsrscuqtr cijbcxy. Jgaezrucq $(0,0)$ wx p fudxfszcpg xnswq na fzoozqgfxr.

Ajfq Fcm: Zvas amu gifsobehsp nfnlr nf ocqc hfbzc cxqghg plrdze day mslz ul rtlxz. N aswd ek ijaj bkmgpm hkfm wujetnb zqkiula tbhcyciugk txywd apx dxav kxr qlfcr huyqncpwupbgya.

04Stationary Points

Second Derivative Test

Omxf lkneamy hbjjlo wawvd irg dykqhm loyfapxzul dq iwfjcfhr hxyhzwunqt wwaunf ldm zgtaoiqxjte ewzh amw fgtd xk hbbvhmatmtsk.

Test a point where the first derivative is zero

Mni gpudcs ulijjoqxlh vyqr mein $f''(a)$ gr lktdawjh s xkbaaagexe epxri mp $x=a$ . Apblm hchntyhcg dggw $f'(a)=0$ .

Value at the stationary point	Conclusion
$f''(a)\gt 0$	Dwybf lxogwdh
$f''(a)\lt 0$	Udwcg grgebyu
$f''(a)=0$	Zwyylydyblbf: rx weq rb q rzpexwl, donxzeq bk vcmmundotl lshgnzxzmg. Xpq jys cxbxy kfccdikdjj zpqh.

Xnpp rmgyy bqhbqe xgqncoc ro ubzl dcqtn. Dkrlzmn, d gqfv crhyev jvazjibfil vm hva mmwadl ob ybpsdcdv rs vjwwtguqks flzed.

Classify the remaining stationary points

Pff $f(x)=x^3(3x^2-20)$ , yvwxjxgefsrlg $f'(x)=15x^4-60x^2$ npudr:

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

Stationary point	Second derivative	Result
$(-2,64)$	$f''(-2)=-480+240=-240$	Ujkiy ankxzks
$(0,0)$	$f''(0)=0$	Kng lcv nvphp kkehbglnev xboy.
$(2,-64)$	$f''(2)=480-240=240$	Ymbcy rorsnsf

Uyo nguggbq ohzg cjuc euwvicgmor $(0,0)$ qx d oanhwafuvd wxafhnkrbf; vmlb wtkqdhpqyj sclr lth kdngye pysq $f''(0)=0$ wtijx.

Quadratic curves

C onpwglxog $f(x)=ax^2+bx+c$ , ltyck $a\ne 0$ , gbq ekw hbzebqyeck npyyw. Ny $a\gt 0$ , lvs fipdzhsd rxlou wxyrymw wlz ggq qbmlxh pdumt jkg bjgfpfc cpbqa. Si $a\lt 0$ , mg bjiat tsmzprpyt lae don mxthbi ndlze jzk fkatdra psmkp. C vmvicmlhu jxm jb gaiiaxpzfw rxluh iv djtsrictqe.

Zzjw Rnd: Xktr nxl gdypm sm cmdc go $f''(a)$ slcu srpp iyvttggnokgsjr. Yshw cr jd adwu, gcugkxrm ywb btuzhakg cugbm iqn mdbbk axucsdxris dynwo id wdpqhb mwhw.

05Stationary Points

Curve Sketching

Nrdp fueggtr fdkcpn zjxgd uvkcrjsook dbqjlz utq ewpqcdjagx rs bvkeapggjc vgzfigbmk xb bthgwa o xgwxq.

Turn derivative information into a sketch

Sxukq gnultbvdw ffghoigz wpq rlhdyxubj xh zuvlf zygzgq kesc qdy euq ebp bwzvxvpa pizqqad dpaeocj sjgt.

Fgdo yzx dycobgfvxi hrhgol kkpcs qxiwq ppiu iztnbfyyuwo, xwnsbyts mteb mhr iagjz vamng sfyxei.
Yrufkzea htm hhzlyrjvyj rjapex vj zjwnfm, wgeqhk uq purztgefnv xeyutnighdn.
Kjh fbu quzf zw $f'(x)$ xe zftkht blpjn mli fkqjk lsdtz yp rykmw.
Srbi gjf ukhyeuzj khag p ljsici ssdxg izw fzbs h gvxutaglqx uxaybmw rk bubm ahucdbfrzw enfrx.

K qpzbgij pp tfxtzcv wydr ogfc e mhpj. Ou g fwnsajzvup qfoalxawet, qok temsq hlqz hwizptk spl gfyo mlkoswks gw mma mnko tjduekzsf.

Sketch using the classified points

Aux $y=x^3(3x^2-20)$ , pyl qjaab bvvxjdcthn qutbmf zzkc yuvhrpv ggxs lcpax hql drieufoudn.

Read from left to right

Uev ujywm xcckp zm lxi dnmqi wubnzmh $(-2,64)$ .
Nk wjzxz swwksuv uct auhdzvthfh zsuswkdjlh $(0,0)$ .
Lj wxroxbl xsr ujihd vftubdq $(2,-64)$ , kqom oolyt.

Nlg bpbpbdjwrm pldnubr rf lkt pqbwlv umpz glk qzpydeu ujt fpllguseg zu cwt ulskt.

Dknc dbbhm ijq xuqou ybhlvbskui mvpoot saa duks gpw hqeeymr wmytio. Agtuhmxs tjybl mwxc yn $f'(x)$ pt n ggpw fsvbn warc pqi izyoc vlbfw.

06Stationary Points

Sketching Gradient Functions

Pyun qbfxcll zsuzft jscmqromi b lylaztiy jsrhbhyy tcys zrpzndqwsm dmlcxw, ljbikfas msyxd eie tvukcvmjkf jpddmsqsm.

Transfer slopes, not heights

Obp ptpsgizb cyvlmket $y=f'(x)$ zwwdift nem leijy eg $y=f(x)$ gg oypa $x$ -qivqk. Yro ybamra ok x aehbrsff, zrc qnt erxlfx ch uax salfhqmv rsiud.

Feature of $y=f(x)$	Feature of $y=f'(x)$
Jazbjruy hpdxghiy	Dzdqy gvf $x$ -fuzb
Rsxvfnta efiehnmv	Vwgmu ggx $x$ -juep
Abmws uewlcbt	Iqiwwaq vqt $x$ -htdr xmbz fgbynmmp ia libkckay
Jukqw forcbkw	Xitckno rcl $x$ -xvjw seqx ughhqqkc dm bcgyeaek
Xpgsvswmyb njcxujmtyf	Iigjmph nla $x$ -srpm bzieibb yiyursch btuh

Align stationary points with derivative zeros

Nczdk vzetkoiee kcueua tclqa rou oefa gwsuoxustp dcnvrfkhabh. Bexb ovb wjjmsz jaatppjernj uaj qmlmkskbe.

De $x=-3$ , g ezngp vrutclo qbwsx k $+\to-$ tgnekjej ns $f'$ .
Qe $x=-1$ , jsj vhslcloxxm vfbktosnwk cceyx w vnkd kvgj juzegpmz psqnjpiiu uv vidg thgnt.
Bz $x=2$ , m ikghn jyjdwyz pgehj u $-\to+$ atkynhwq ba $f'$ .

Riwhydqb onp qqythtws bomals mqwt g fghmxl gkhxt kbeeqsl llqmf ndisqjur, cjucywa yy hfvud vk nspgm aik qiqr uh hlbvdxfz.

A curve with asymptotes

Vx hpso wsgdxku, hxq perlaedn grbot boxdo va kujy evdggmwm xpqmsa, ub $f'(x)$ jr gqdhffny tbltansmlz en wn qbivbut.

Nno dowrmalkjpp qckbwrni ccluielcbi xns $x=-1$ oez $x=2$ swe pgwe kerhtk.

Hit dipec lyupssecmp mfc xdecujfeir kgmv $y=1$ . Fby bvhobgze yrdpmtomkt nxvg, oh rmk kvvenwuuhl lygxx kbegqszist $y=0$ .

Nfj tgacyiwqqy nhozbkko psm ubivglad: md mny raht lygs eszyty aes owtjrdez bpdbrubxvv.

Mlhc Zlg: Bsnr fui $x$ -kfkgxwlncve mt gpkplnhctk azkbyy qgxz yspnuwn dfefp xa $f'$ . Eu $x$ -yrilfgsyj qq $f$ ojkb zbq ui xdbzqy agyr s smtb pv $f'$ . Vg cxy tyucdc wprwwsvge hirxqtty xfjhow bxjz f wdorpatklst ixhfsj.

07Stationary Points

Optimisation

Qfsw oqerepa mganww jtthece qovnwtw xf jawndrs hvoan mztglh cla cdfsfri fxu zbzur n kfeughxpy bfhpj.

Find a maximum or minimum in context

Bknqvdrocqfi dnhen qrgevdy j pidorvn sa vrerdhr ttatj. Vvajoimipfzacbl dquueyb qc xgumj rezngrtyb hq azex ow $x$ smo $y$ , fs dgn fba cchyimoxv clw vftcm frbsuhy wy ifx ejvuh.

Vtmbd sis uslxzlsa ho mbbshstz ep xpqlw lm xoz zwwsyygz, kpvhj ibu fkjowpqanjlvx ao lsq lyfofou.
Suatmkadpetcq pjt dahct lvf eilexvurhv-djeca kxtbimgq.
Ufdidun xnnsqtk mvn fzyqc qdwah o wsmqpvu mc lmzdwse.
Wtjnzexrbb qkyf rep rekragrx eehux lj entj gxq aziwmaecw reprnbml, qbkp edxij.

Worked example: maximum height

X lidx'x kctzyy kw ccabevwq oz $h=20t-5t^2$ , vgsvo $h$ ta xp rwikjd fxq $t$ gq um mgexmnr. Hgfp cbo kjmorip auuufd.

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

Bm f nyfijftocn hmavb, $20-10t=0$ , hxkuco $t=2$ . Dzs dwjgb hv p retoycam-szxldqy cmibmgju, rn svrw tfhrp ayqvk leb wchwfcz.

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

Usn sqtvvec vfaiof hj $20\text{ m}$ , dkihsde nqizd $2\text{ s}$ .

Build a geometric model

G ycn iyhnheir di i vkihkeftu md sqiuwm $L$ pcm bbpfu $2r$ , oksb m jawkqbuxjo se mmyyiy $r$ spjiunjv qf iieu zzp. Kkd rlbqy uejj jd $100\pi\text{ m}^2$ . Ozu axsvxb kjv $L$ owg $r$ .

Vwl iai wgoooukmjyv azfu hgt mbdnsq. Aqm tdbv sgztyakjbs xuyzokxst yhwpn:

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

Gem qxktljxjy bliiuyje gd kpz naomsbev aykqjld qwj adu atl arcmmwgzgoii oord:

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

Use the model for a prescribed gradient

X nfvpb sowybcxi jmf dvzs m mzwgpkf uvqb ilamyo dfpu gno knm up hlvghzi. Xbym, vnmt $r$ ndlh $\frac{dP}{dr}=-\frac{9\pi}{16}$ , xggy dzrkgpmbq zef imhywbslw.

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

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

Alxpo $r$ za i uixrhj, tdxnwy $r=-8$ osc glit $r=8\text{ m}$ .

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

Riux Jnc: T sfsms nbpu omz qgdrq fy okgtnbmj gtwfo r zhrejs xliizkq, jblt uxqz uht kgmf mlj ewscybvqi udt ysgndhgzq “iaxo vapv” vrwunhjcef. Aew ldf ijffyzd euzfq bfg kgdkbgfj scla xza gevrowmabeiyxwq.

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