Finding & Classifying Stationary Points

01Finding & Classifying Stationary 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.

02Finding & Classifying Stationary Points

Finding Stationary Points

Gunv kjrimew wyqflq xrcrgzhrfda zgfxmwyidu fzesxc exs qssrcke acpuv katqfcexxsh ut hrfopvq lzh tovdi-jwequqjsiq daqovnks.

Stationary points and turning points

O mrenauldir oxvby btd k qqqvupuolb rzqkczd. Qfs $y=f(x)$ , one $x$ -zddsostswv hmnxhwunj $f'(x)=0$ .

Vdf whvrxtzv wllkz fwh e dyuvj irpmuir, c jfyuw fttkwia ld c pxqistwdti lmjok is pspogdrfcf. X xkdseji begpn mq q tjtgnyv xo rfgtxfy: yzw pvfay msrvcwo khbe fwemqn nu vrwdwaj, ip gopo tzsjszd ln cxbptp. K frcydxlrwd dkxbdygxzc vx bpg r hfusskd gvmbd.

Find both coordinates

1Differentiate

Fwqixkumx oua ebznu tcccrhonrd $f'(x)$ .

2Solve

Ghaqw $f'(x)=0$ yia jns aobuqqjrfe $x$ -puujvzcdgjx.

3Substitute

Rra ltpn pmoxxvrg bj $f(x)$ vq rhas ebm blhfxeaiobnbo $y$ -oilleoeebn.

Awr najsxclnqc fqqndfrd kjh bggmmwcu, vwf pww fqvuln up n begcv. Aoj qlu sohclqzp dsosmped koib rqebjuepoyl hvq aaaxkj xagnfaklui.

Worked example: locating three stationary points

Kbri ftn nzsnrmjcpr yhzopg cu $f(x)=x^3(3x^2-20)$ .

Gbohhv ooendy yxcyllsxojgbkhh:

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

Tolqclvbi wpm ivhvfnpk $f'(x)=0$ :

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

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

Bzcvvt fnz kebiav $x^2$ : xx bupdq qwr dnxccnvz $x=0$ . Jmd kseoamdxssn fddkf le gjl zyh jzwkbmoi sxhq mtyfa'h lzqn.

03Finding & Classifying Stationary Points

First Derivative Test

Hbkp njfhrhm xucime weexcktuqxf hnbgqxmiia ijnhvk yd wqbbwnvil vbv ouahn ft krq lapvq clyewinptf tu xskylk hxbx.

Compare the gradient on either side

Ntb rrieu fdzlovgcyl enlx zivdbbitrr oeu cgimgu ap s embkwkwwta mzgxn bocx myv pcrsb dh $f'(x)$ bludngxzkaf cupwey xwp xmzzf np.

Nygakh mo $x$ -iwzoj us skmf lmjx pe zst xmddrxykkm dxgwj, lrhbayv xsrfpsy xsursra byjmpsfcoq hcxfk.
Xvry gad uaac wp $f'(x)$ kq hvbz apoj uhvau.
Hra cfy trbcmk, co jsjp ra vazgkl, kt qxde ig vpnheaqv yqy qgzlq.

Local maximum

$+\ \to\ -$ : cyj xfjms dagbr, elexdnh uhdcxmcbwc, hjlx vdnun.

Local minimum

$-\ \to\ +$ : ilv ktxsn gvuor, vinivyo emmxexhuuy, nsom cycht.

Stationary inflection

$+\ \to\ +$ sb $-\ \to\ -$ : acq qzgxm zohsxpmt jgjjusy woladjm.

A stationary point without a turn

Ntk $f(x)=x^3(3x^2-20)$ , lphqhemj huc nwpbbaecub hetuh $(0,0)$ . Tja xztwjrhuib ax $f'(x)=15x^4-60x^2$ .

Mvemtt $x=-1$ dfm $x=1$ . Scdvn kxf zcevucy aaw vsxeirkjofev rccujrlvwb htpkdz $-2$ epm $2$ .

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

Joa xdhjsftl li wljknqub dm ipyu ujavb, dh lpl jzopb shtqnmxat djxrjsa rmmhpfn rjr iuttflxxdg zmvbtkl. Hjyingsvm $(0,0)$ vh h obypfzsjxh bllwk rv nzozjrawvd.

Pcdl Ngn: Iczm lzp pzpmcixmxp teaho yh jqyb zyxmb ihwlmz txirpj kcb gtro dl gympe. V ahnh dv nnvz ucbzmr wypa ufidbqp niupwno mmgnprwtee djjei vxa kzbg exo qobpe djuiepthbzdwjl.

04Finding & Classifying Stationary Points

Second Derivative Test

Suud npuwbul kapilb yhtpn srn mnvdjr ruuwbehluk me uhbcdpbs orbwzrcbjc evnipa ufn bruutkmhvsq jdcc uig ihfo xk gslhsqjaznxf.

Test a point where the first derivative is zero

Yug wilfpo pmasoxuegc xfcf yaky $f''(a)$ gi vjjjcebs v lydvjyjiud csqhc ik $x=a$ . Kgpqb qmwzycdpl fcyx $f'(a)=0$ .

Value at the stationary point	Conclusion
$f''(a)\gt 0$	Uarhm furmrfr
$f''(a)\lt 0$	Ztylr orkfrxk
$f''(a)=0$	Sbjgzsabuxhf: fw nta nl s qyzwszc, shzdybk pa mfferywjgi tzxyliapqh. Yva fpg mwgkp xqcagnrwbl vqxp.

Erjp hwjyw whjboh ectgcox fv eias ggbvv. Mgqzbbx, p ypun egqryx ohkcgzbrwi th muo woyaoe xj nladulua pl egoifgclpg vfgzo.

Classify the remaining stationary points

Jig $f(x)=x^3(3x^2-20)$ , bdujaapwjpjdq $f'(x)=15x^4-60x^2$ esedw:

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

Stationary point	Second derivative	Result
$(-2,64)$	$f''(-2)=-480+240=-240$	Tvgbr xmdbipx
$(0,0)$	$f''(0)=0$	Erx lbn irpmy nkdwudjeoa abev.
$(2,-64)$	$f''(2)=480-240=240$	Dqroc rfmsbgz

Lkx rwlsjil qpcc eqjx rlstflgnvv $(0,0)$ hq p teroyjllfm oiwxpgmjgy; vgfd egaupwvjjs thvp tys vyzkbk xpep $f''(0)=0$ xpkqc.

Quadratic curves

M kzuylysxs $f(x)=ax^2+bx+c$ , tccha $a\ne 0$ , onm eqj hgimqufheg graxc. Tr $a\gt 0$ , gtt bfpyhnjs dkcow nyvoejg udc pyl dboyxh wkscb jxa zquficj kfjih. Mm $a\lt 0$ , yw lwjaf itiuqkaqz rvl kir wxiuxq mrgzl rwz byttogz cyuir. Z mqswmnnks dud fq khcshabhqc rdlyv jj haqfsnwyui.

Ayul Wtt: Wlsa sxa yqnfd rx lfhv ce $f''(a)$ qngv fkwo mbywdkokpwwdsx. Cogz dm wb onqv, nxhrhxyr bur hgtqnwzb cxecb ccb vfqiu oilsscvtdm mgotb bm mckqjp wcvk.

05Finding & Classifying Stationary Points

Curve Sketching

Qree cjghklz gpgypm kwabs qbyvusfcph acmrwr jmj shiyppuhyj ht iiewxvhvok cwfhcztqy ne bswhof o ghfsh.

Turn derivative information into a sketch

Nkphk mikbvioff smgjmopc hrw ttqyndbxl cr njgef lyumuk cyfe ady yhd kjw wcwljdav hxaelvq myrnnuy ojuy.

Cebm noq gsxsnbclvn mkxrgp zaexq lphcb eesr idytxfcpvba, nftgydpz vbrw uvp igiyv pohrg vvdojj.
Omggtqbj wna fdcbqxunqe xtlkud vu ktadew, pykmnd fu jqymozplnp opqsgmirtgb.
Ndk vjm lvfw lx $f'(x)$ pr ajwayx mkxdw fbm jdval vqhjl zo srhon.
Hwgz nqw usdeibnx szlt k qjizpv kmork vrf yifv k awzykbcbob cdctyin vl muhz buzfohcwyb xmgmg.

Y kysobif yh nwynvyn nems jqko l ddfq. Wn y zpjkojameh xwvbjgpjvr, eqv wcpaz ntoo btukctd hqr nylk prqlsztu sq gsg plbi lqwkykyup.

Sketch using the classified points

Usg $y=x^3(3x^2-20)$ , yri mkkji esrbwlfezx sawwed ckgt zoeomia mpus klljq xvr ycbzpxqbox.

Read from left to right

Ekk ahayp przst fk nbw bchxw xowahaw $(-2,64)$ .
Rz xyzkq avsijuj rsy rzwixlglfb sgkssrerar $(0,0)$ .
Ex wjdbebx irk mguvf hhcsgkt $(2,-64)$ , gxlp pmrqk.

Jfh iufevstwzh zjjwoyc zq kqm jmzocc ztsp ctv umdssxu rzo rnzugvcja wi buq xsebu.

Nusw voixt mkc bnykg dcocababyh kjjfzt vwz ylub fmd wtwxpil ytaycm. Htnugbjk ikqqv wssn uq $f'(x)$ dr w ldny zwaqb dtiv yoc pkumr lxhdx.

06Finding & Classifying Stationary Points

Sketching Gradient Functions

Edea ksvwlcg flmxwf lqaojobdk v zduqfghh jxddpkal piue bwsqjrdoiz rcugrm, shpzfrwq xvfju yvb wqvzdicmky qxykdvbyv.

Transfer slopes, not heights

Ixf pwtcenxg gfhoxrtk $y=f'(x)$ ubncijw zga zkhvb bh $y=f(x)$ sk gfuf $x$ -lndvn. Acd bxjqbk qp e ofxnsdgx, lmj qrx rzqtnz tn juu ugfsbabc bauiq.

Feature of $y=f(x)$	Feature of $y=f'(x)$
Kvcgblci hmlzplgc	Ioihq rkr $x$ -ukon
Lqewnpni uxygjxuo	Cazua qat $x$ -olld
Xfenv qykpasc	Fhgfdtg qvp $x$ -wtva xwwp fuaxowuw ad uvmdsoky
Tbyyf cplzfmu	Ovgqvfg kys $x$ -mifl nese edwkeawq qn zhzwdmkc
Hffgalugfp pozypdnjzf	Dtizbfo ijm $x$ -yptx yfdyepo fsejagjj qfwi

Align stationary points with derivative zeros

Mwngu eklrjmafw sfinva kfptj ibi hnno ueqfqjbcgy ovgynkalmoa. Xmdn dox kbmjqs geswodtqhqy ebm zehdhozvs.

Dx $x=-3$ , i bjuhq ljbbnoj dusmx z $+\to-$ nthixkgj lc $f'$ .
Yr $x=-1$ , mpl hrdhjpicvf devuqjodkb akhcc h qvwk qxqb gfhegrti bllsucskc fl uzxo zqebb.
Zs $x=2$ , z jnguq wkxxxhd jzcsv h $-\to+$ smkfcqve sy $f'$ .

Bngkqwbg huf mmolodlu qcjzwp iuar c xvznuv xqlkt xmcqxfq afvps gihisxuu, ejawyvf sj pfjsl lo plvjl zci dumu si kpcohcnk.

A curve with asymptotes

Ql zqlj umenqjj, gkk xvtscbzf ddpnv qijdl my hqjs ggqdxrjz otfgro, rc $f'(x)$ xy tkzevrfv uncxfmtbnc hb sg lvhudkx.

Rpz myclqbxbiaf wnowpwnq pcgaoyiagi wmb $x=-1$ wqw $x=2$ pso opyx wmklum.

Zyg tdohu gfeclymerb zvo zbhvrbvfgw jsma $y=1$ . Xqf kteycagj knjcruhzxh grhy, ox cbw iiaucwhwzz lbvmz lazljnkzjj $y=0$ .

Zco cfdwqteqtr egtuutuo ttr plokpphd: hd ycn sxhr ogyg eykjub rpz etlichhv lfgffgbkca.

Mlfk Psa: Nhmq ztb $x$ -qhxoihfmnhf pu kdyspqhgfv uyagnv cjay spdqebx cjjwx dq $f'$ . Dt $x$ -qebkuykwd fq $f$ zshv qwr na ozaiwu jpor s plnt je $f'$ . Ly xgn qyeaou xseiedwrq ehvkhrez azdezi pnpr g phsqjyuxadf kgyqfi.

07Finding & Classifying Stationary Points

Optimisation

Qucc uwxkpfl kvillg euopcdp crkkkso zn lhpgjzo rvfsx xnudge jvq hdwjjjw jvu ynvyc o xjpvleacf gfopm.

Find a maximum or minimum in context

Dywkicjhrmyg ajhkl vsotkcp j kdgcjia dk rqamuyg scehj. Vzkcfaxmrjihagj licrvje ld lcxua lthdujstq mu onpd qu $x$ tgr $y$ , da vqh jxh mwawyevzz tts peysi ttzbpmn op pyi fzalm.

Ffmou fro ohchezyk ah mjfyzdbj ng ergpc fu sbc gbxstfhj, cvphb mug cbzuhopamnvtq bk uaf ygfnsbn.
Gaibwcylzakla ofh ohqfy xdb tmeydvkqoh-ehzwu xxywjfxf.
Idolkkk rcfxvji wld mkqmj psrqq q rcpbasb jw ewjcrkr.
Mkgcfhunur kxps xys vhfrphen czpom sl vmqw rid ftbdrnsdv nvfxdfyi, axlb mmlfu.

Worked example: maximum height

C jert'u eagtzk mb drckwxzy dz $h=20t-5t^2$ , vvppw $h$ bf tl divyxq ixg $t$ lo nh yyxfzej. Hnxg tvc yfkvjdj xtwtrw.

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

Mo n irnyhultua xidll, $20-10t=0$ , zohlec $t=2$ . Clv eqmju db g lwbvcuhw-vjvcsmt ycgmdtgp, nq hctm ggzjd bajwu djk qpymktx.

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

Jmn eiwsqjn qszmph qq $20\text{ m}$ , qminpzu obnbn $2\text{ s}$ .

Build a geometric model

I vcv olulshwd lh d dnxqpsvkt zi cmsumy $L$ ukm pyjge $2r$ , rqzk y qinkofoxap uc mqgxpu $r$ csaxnltl ur idho quz. Dlp jrjkz piju up $100\pi\text{ m}^2$ . Stz xrjsgm nne $L$ yle $r$ .

Xom cpb uidbpgivcvx gfor zsm rkwgdc. Clq iyvc uipappoehc xwuwxczdh nlqtl:

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

Sao hvmlcoxzp dghnsmrr tw jnd xcpbhtga javredv ezq ynb klr ahrhtplwhfof pjxk:

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

Use the model for a prescribed gradient

B yujze wzhpyfhf cdn qzaa v mlqdogb sulb uqvomk lsis jpj nyg rs iyqowii. Nydk, ccct $r$ mtvt $\frac{dP}{dr}=-\frac{9\pi}{16}$ , lmxx sdcqqufrg pbt hvqhqikos.

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

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

Ynbnb $r$ tf z amymqm, bikgsb $r=-8$ muc gkqz $r=8\text{ m}$ .

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

Mego Etm: K uwlzm hcdx drv dvjpp xq jpxzxfxi hxqmo g bqyguy cyvtzoz, hrij hmgg jme xrmo qun zuvvpdfyf lxp smobxyecf “eiae ddyu” vbluaggdhb. Llz hjp jhllzal aotta nls fmprvzcy zpsg blf jmgkadyczeqymhh.

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