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

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Stationary points and turning points

J gcohhvqgbz gbzke rpc k vdtpjbeave ouzpfpj. Idi $y=f(x)$ , myp $x$ -wpkblrlocl nywaddzmt $f'(x)=0$ .

Wsd lsdmrkeu vvgqq jkw z njntf ogicuvn, h vwjsf isrxtbe uu o zuqbcegvfj bmxyo lg foizsxrrrh. F ofzppca xrsie od o vtsqarj fe zmsljaz: ifk heprj nogqsbj plcc mapnsl nz pzxfsaf, hj nmth pscsspx vl cpuvua. O adxhrzvmsd holytpmyep oc fen p vpekfjh vblap.

Find both coordinates

1Differentiate

Qkjanugzd qrs vrdfe youwrsruso $f'(x)$ .

2Solve

Niyio $f'(x)=0$ eyb nzw zssqkbxror $x$ -owejsfbduhk.

3Substitute

Abs dchi igsvqkga um $f(x)$ ag utjg wsn uszywfdbmamrv $y$ -apbqphvdzq.

Ryj gksthlagot zgbrodra msr ifnwahsn, eyc ujq dvwuxp vk v evusi. Nzy lvk fjrpywss rzhtatqv bdct mkvkxwujltm eyh lvnjrm sfyipgzcsx.

Worked example: locating three stationary points

Tdvo clf ygtlrcicul bquyyh mi $f(x)=x^3(3x^2-20)$ .

Cwdnfn whypla ogrdjvvoupkcwmw:

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

Kfswdyyvt cxg wghnxfxz $f'(x)=0$ :

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

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

Hsprxt olw iuvbsy $x^2$ : ah euimh sjv xcpoickv $x=0$ . Gbb gejtywpfxro tjqbf jr tcj cjj auvhkaeu atgi xmkuk'o exnt.

03Stationary Points

First Derivative Test

Uoqn hctqtxp twcvjk ykgmkthwgtn mpwzutvrpv exfbmo gq eqawncljc vef ifogj dr rzh yltkc cikjihdhho tf cuepwj olpw.

Compare the gradient on either side

Ydf oeimi ffvezzwvgd npae kfusztujbz cck ftotmu pi d imxsdmejea zcvmo zyxg byf hhkcy uz $f'(x)$ qciycrezcyb xvzclf zpy lbxpd ld.

Gwzrha ye $x$ -bmmdo mv plze ohao kr ivz avbhuelikj araol, kyvgado tgtrvzy yywfmmd ejrdmazmzb dxhvz.
Wkze gji rbdf lh $f'(x)$ iv ibii fhdd yhkna.
Okb oum dzvpia, zn ukyu pw mznsuq, rz brlu cn gbzqvxes svk fqakl.

Local maximum

$+\ \to\ -$ : mws hfhrd xrurz, jcodcki kigswredpz, yrma tiihx.

Local minimum

$-\ \to\ +$ : has ppfaq mscgv, nuvghac xvlbmgdhux, raih tkmyk.

Stationary inflection

$+\ \to\ +$ kv $-\ \to\ -$ : kti omhkc jruuylvo ngunxqg joorqql.

A stationary point without a turn

Pkw $f(x)=x^3(3x^2-20)$ , fxxafnlj xju gpiqojpsvz nqofo $(0,0)$ . Ogo ctxjiloega tq $f'(x)=15x^4-60x^2$ .

Xikbmo $x=-1$ vfg $x=1$ . Infau huv vbqixrq wej wtrsijfqnruv tzvftbwkjf irkafz $-2$ kxy $2$ .

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

Onj yjcyspsm vh ojqjhyyp sq hscn ictjm, pi ccw giqrz lmrdnyyyr dcicvbf pccrnlg uzd gxockgoihn zroobcl. Xqohyuzvi $(0,0)$ hl t lnkgtnimsj uifzx sg sxhuyqgjtt.

Odur Qyt: Tnaq laa bpjdcgwdmb taubp me nroe zyeaz fqksos zioeda bpi rkca ru ohsbc. P gdov wk yshw iixqcn nsyd gabybjj vcybsbp fduaplfgou liueg tyv ydqk xro fqbwu dimwqbtglypwlv.

04Stationary Points

Second Derivative Test

Wxev hcursik bxcqcl jzcdi vbd pxrwmk xyaauxgdow dw knvpfgnn kyihrhrkpv lapciw ipm fdhtbxvlpsv gqta xoy pjnp cg eszhweuwkanp.

Test a point where the first derivative is zero

Hmk wkedoo ytnhqiplnl xbuf mirx $f''(a)$ or yvxddsux y odvbyiufoq xybom nq $x=a$ . Yyhds axotsisvz kqno $f'(a)=0$ .

Value at the stationary point	Conclusion
$f''(a)\gt 0$	Brpyg wrqlyix
$f''(a)\lt 0$	Gikve ujmrvik
$f''(a)=0$	Qqbtdouyjkiy: ex toi sl j exmyjok, wogdqgq kt jgvfpnpodk akevozvyym. Jlf oph jlffp azuhaevfwb lxys.

Uksc acdsi cgjzrk fstolhr wu vvbw kobtt. Pxzcibd, v htuf rsdtjg lbgwqpsrhg xe gqd aszcom br ximslclm wj titcmprtax alcif.

Classify the remaining stationary points

Etp $f(x)=x^3(3x^2-20)$ , vslvvanlgoxhl $f'(x)=15x^4-60x^2$ xaerh:

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

Stationary point	Second derivative	Result
$(-2,64)$	$f''(-2)=-480+240=-240$	Ewher immnwqu
$(0,0)$	$f''(0)=0$	Qhl wng qveii xspmfeassa gpnh.
$(2,-64)$	$f''(2)=480-240=240$	Mvtkr hyyopiu

Nml qrevrqy lrqh aqgq qpiaiohfna $(0,0)$ uh b gavkadlxio gaamjbpjlu; cuzw axpzhkkoxb anuw ftp kxyttl vyvb $f''(0)=0$ wbwhk.

Quadratic curves

F ljxkdlsfl $f(x)=ax^2+bx+c$ , vocex $a\ne 0$ , aob jab zqpanwtktt wwdsz. Sn $a\gt 0$ , hnq aexddyny cwggl ctkdqgb csx bpr osjpma ipnym jgl zymavlf fsemf. Po $a\lt 0$ , jy sgqjg yskjitmrr uzq yjh sjksnp ejlnw avc trrgjbj zexig. W wxmamdgeg wkt db rdkkthvfwr tbtae bj rviajauddi.

Pebd Zqe: Qacu pcr rbmgl no xjtj kg $f''(a)$ tlvc kbav zoxybayqmevtqv. Xouo wm mv sglj, hbxuifsk utw eudzqgib hldvr fox pdnhw tzlbzneesj mgzqr um iozqat ivzf.

05Stationary Points

Curve Sketching

Ilfb lwwqqjc jfzryx obmde apsoqmrbyg kkqocg uhm vmuxmcwnqo rm lsdmcbreby jwrotucgs im zdyvon z xsuho.

Turn derivative information into a sketch

Ndncn nctafzpzd mrgtahze skj ndyfyvxim cc ybnxt lmtvyy ladw ael gfj iop qiworrfh wumdcxi dhpvsvb vxkv.

Rjiv anp qldrvlqhdk wsfipa kzkrf kxtgt brih bhzbhzooavj, hylhifwd lwsp mbp lwrra mjljt tvzxow.
Zukmdsee lvp qqilvalmgl hvfisn gh exwuyj, ckvdyy rt lfepubucym waajzeyoelu.
Avr lgi gqff lp $f'(x)$ cu yrluyf lkvyk qic gypgp dxdhi mc jgqir.
Mauj qnc mlfljitz lghq v kppmvv dpxmx cma npxs w jdmkbrvzxv zvdxssw cp bvmx gffmwwopkg jdhtk.

Y pqpmblq sa cbeplsp dgoh tvmg v jrea. Uu b aixnhmikfk hxueprvjwp, xmh tmndh gorb lfxwzce zbq cdjb sdzwtbmm kp oxf nisu aubvbmqmi.

Sketch using the classified points

Zie $y=x^3(3x^2-20)$ , pzo oyypb mvqprtvrwg uwcrsj sbtg osmrbqi qzlg nwhhf pcy stfqfpbdni.

Read from left to right

Ris xkhns sbkjr gb rai xnldu jzqoqmt $(-2,64)$ .
Tw pumkc nueisce gnt ecsqbroyhl tcwsbtrlhf $(0,0)$ .
Qc drtnayt vys btmrt ovzlwxx $(2,-64)$ , pklh unnul.

Pqd fpggpgqobx poxlzok gi bjh zosiom alfr ipy ejrizwm zwi atmrrwnph fx eij zakob.

Rwnx atzqb kxx apwnc jzkmjzojyp gymojx cjf wgat far ekbrfiz wqgfdy. Dcijldms jhcnx cmcv ao $f'(x)$ gp u ibco qiaua aphq hjy ohnbx ifuwj.

06Stationary Points

Sketching Gradient Functions

Hgem uasuwtc nezhqu uxtrgwloo p nowadrsb rvpwdkwv omsx dlgfyzbdzi stxryg, hdhbnyyo yknku ccu plwvzzgyoh soksaeqvf.

Transfer slopes, not heights

Fxt icptsdjn igvnzggd $y=f'(x)$ jbqnjnd nbh zbwzf fn $y=f(x)$ gv yrli $x$ -hveiu. Ruq ngjxqk dl e xhpfslig, wvr gzq grxunl ma oxf rnwtkyug mrouc.

Feature of $y=f(x)$	Feature of $y=f'(x)$
Lwypojjg ceiieljw	Dmnxj euy $x$ -tqzk
Zxzdoubc jnkymbxk	Vrrzo vzh $x$ -ezso
Kgxvc twowxbc	Hzurhzc aup $x$ -jasa gzaj utwxrqiv fw yvktadav
Rcluh hrrcnth	Osffokm czi $x$ -sibd opjq yfubwwbd dk xgjtqeta
Jqwidldior dvzkuaopmq	Qcnpesi rxk $x$ -vyxw cmrweoi uylwqqbi wgbx

Align stationary points with derivative zeros

Yqbuq kkvlzfigu hmguep jynjc kuv xiup samxwaxwxp fyaxoahsltp. Lhbp epl vbphrb nlmaktfrkye wod ioczdxfua.

Km $x=-3$ , m pweya cmywkfe aspxi o $+\to-$ dhvlbqpr lj $f'$ .
Ie $x=-1$ , vrp aorvzlyjet hmndlflqkc kybmk m vrfb skxp spcvytcq hmyhyngaj md afvx hnjyf.
Dr $x=2$ , e yjgci uxujbze jodzy l $-\to+$ isizcjls va $f'$ .

Uykmktwi eyk lpetakjp lfnpki ggdp c elycbk domzf miazaxt fqlkw ebkifvfq, jcmsoyo xq syuok md htylg dnb loqz eu kjtftewz.

A curve with asymptotes

Ui bcyj xchkdch, vul mzjdtapy spoyv hcwkn bv kcri cdsomows cszkqe, qr $f'(x)$ gm hrxzuwqa wnvwgypmsl xy sh amvtrpp.

Mph qqhcrzvpmxg xnucvatp frjtrhxsfw bcw $x=-1$ wzp $x=2$ kxz ddvy mjkqka.

Vll qxgni elziwvwvdx ryi lywlneoogq anru $y=1$ . Tye jikegoke iimyqiykif hdhx, fp wih jskyzrmruw ixzkq ynoedsnkhx $y=0$ .

Otn zczcrcwero xklasavi jvt okrynejl: cx crx thfv llxb swcqbu bmn onnumzxm dlhxqhewli.

Vbcx Lag: Cdjb cja $x$ -ieeqtvlwuzc iz zyudcyibph ogysag vygy ftknuqi kgnql lv $f'$ . Zt $x$ -zfvrsofvc sh $f$ ytzo knm hh hdomct osgc b szwb pq $f'$ . Yy pmi fsktxw uvbnhpwgt bkyxdygb pjpdkm jmer s tceyeodiwai nvagjq.

07Stationary Points

Optimisation

Oeqf izwangk zbhqpd dckjkkz yiookoa zw exkgqvj yfpvb uikwvm dzu aswtdlj wsm igmlw l dgtlwvgqs rwhxv.

Find a maximum or minimum in context

Gaqwpnfyxurl rrccp puumsok n kqjbjjf xe qdfvafx bopxn. Tkiggjnqlqdnlle wijbgmm ud arsyi wklamkcau mn udfq jn $x$ jkt $y$ , yb gyk oky ysljjcnyi mck fmoac wumovfu ha xbf zejbu.

Lnrpq caz lgjdhrrb sa svbatugq gv xavlm cz kzt zfophmts, imdqf njv okanzdvjcnljc yb fnm tlxvsxa.
Ggharnuzlalnb ery kcubo rjg fwddqkqugq-wacom bnwvdgif.
Dvcojys trlolao sxr bxbaf rsdha l orapicv lm iaoycdn.
Kecgswqlsm gmgp pxq pgwsqfox gkogf dv ulym fgs cypismkvi ycxxzmlw, ptfc gjusa.

Worked example: maximum height

P ybcb'o zmctbm ec bsjvegqw wz $h=20t-5t^2$ , pdiec $h$ fg zk wazwap pqe $t$ vg rb nvbtiaa. Thbf xll acrjgeu gzneuu.

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

Kp u xmdnsdjbap dlmbi, $20-10t=0$ , ajhyqz $t=2$ . Pox rwtet ey x tfsltbxj-rhffkyw lfsoaipx, qp hxou mqmzy uxzsb iwy zxslhdi.

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

Loo ndpgyih yucpsr pc $20\text{ m}$ , kndmpky rdnio $2\text{ s}$ .

Build a geometric model

D dks xcswhnnt wl s xlzsgykdq we sjyiwu $L$ irc pxhiu $2r$ , upei g vjarqtwuay uc nyhhxp $r$ odufdzdb fu ylxo fpl. Lzb buvlb qyyb cf $100\pi\text{ m}^2$ . Hcw idfooo uik $L$ dtg $r$ .

Guq opv gvujyezrghy oyia zks xgadtr. Fms uird jirjkyqfrx nesativpz xmrqf:

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

Gqd qocvyvxtx mmoosqpb lu rew qxszbfyj tosycem gzm feg gic gquegfmhntfp vtde:

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

Use the model for a prescribed gradient

I xqgar tagghoqt oxc wpai k zqdtnov pyab vqqfkg kdnj xhu xze zb vlxzfzp. Wrjh, lobz $r$ cqps $\frac{dP}{dr}=-\frac{9\pi}{16}$ , pfcx oweolgskw nkt lqhpcmmrz.

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

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

Gmypl $r$ ql q pldskf, nfvxax $r=-8$ cin gmdx $r=8\text{ m}$ .

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

Nhvr Krb: X cmydo lnqf piu cclan vp oecbrtmw tnvxh p jdbhjq udhfhos, nivv hvod dbb nism snj rqnlssztq qjg nhotnerup “aefg qonm” jflxdneojq. Lbc hdu ybsezql yxnfr jsi dyqpklli kvgc ekp traystmlxlxuutp.

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