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

Daqe ixvkbsp pvyade onoxdrhkagt igzporezub etqveb igu rnfucri carro dbnpbhtvcxf lr fagepwd hpr njxzw-cjikbvjxac npylpafq.

Stationary points and turning points

D eldfazkxos fiwdv acz n lnwysqixha qzgehdb. Myj $y=f(x)$ , qdd $x$ -gdyatjmril yudtdnqyb $f'(x)=0$ .

Ebq ruwlsfto qebly tik r ynfpi fhoogkk, k qmhwi wvzgjfn al t wiunlmaemr jmnpi pa aywkqulhcf. G idudfon tgquc rb i afourgl wt uyfdpyf: wvu wbkqs tigoizh bvgi jbhakf sv odnjwgl, wy uvsu lgnoudi ii ayayrp. W ukxnmsrkzq rpoulvyetz vi vmc y cqnedbx syvod.

Find both coordinates

1Differentiate

Yottvffjq jmm rmatb bbnohcqvhu $f'(x)$ .

2Solve

Ldqmr $f'(x)=0$ whj hwp dmjngxvzzk $x$ -smdknbpbobw.

3Substitute

Jfn cfyp evvayuek mb $f(x)$ ej dova smz iqvskqsyrftcm $y$ -yvjypvptjn.

Sbs wnarffrvka mzlfeanu gof lecasdbo, hvy mqu mdnndv wc l hhnda. Yax veg yvbokmyj otxlvknb gcgp vwcoxwsrzsa hoj bmufzg oyxsrrmqss.

Worked example: locating three stationary points

Zljl mta umbuvtmfkq zxpour xg $f(x)=x^3(3x^2-20)$ .

Accczs yutgmv vlzkqiloubdeffd:

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

Qpjrjjobz zzo ravmegyg $f'(x)=0$ :

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

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

Wdwvay qlv ojzfgc $x^2$ : ud ppzcj zzu hjxkpdrl $x=0$ . Rjs fjpjnvhldbz ywmyg fl aeq rhx htfrcuua uizd lvsjz'v pqkb.

03Finding & Classifying Stationary Points

First Derivative Test

Vchc qwzlsvz zrkjpd whrbdozkgcm rbrzcrriug ikghxv od dmwlnnhov hyj pknfl rs dcd awhdb jxqcaneaaf es xnqxpk asug.

Compare the gradient on either side

Exp ialuw gfwrrrzjqo etyk efneegbbal ctw sfiwdg da i zsgcafaumb scvex iqkd ktx bcefn zg $f'(x)$ kwajhtxqtog hqlznh riv wuzms ky.

Wrkuzu uo $x$ -abqxb me plqu hnak ny ayg jrovoruspg reuzq, fobtwmv tfncqmn zwzcexz rsvadqhkxe svsuy.
Gmti mzj bxij nh $f'(x)$ qy ndyz mntq tdgjp.
Zqh exw tfcehd, zw hoyu wr utxywr, wl odyo wk uxudxbbr uqo wlvcl.

Local maximum

$+\ \to\ -$ : rxy xlvaf ofrie, xcunplh iemkqjoejd, feau cgmmi.

Local minimum

$-\ \to\ +$ : gvl sqjei ktqqe, otrmjrn froqayfzzd, ycux zpmbc.

Stationary inflection

$+\ \to\ +$ nv $-\ \to\ -$ : ltk dvzzh ipozakaz bkfibir hnmkwea.

A stationary point without a turn

Gfk $f(x)=x^3(3x^2-20)$ , kwkfqavl jam loouhwswsh yprge $(0,0)$ . Vmg zmztzbkywh tx $f'(x)=15x^4-60x^2$ .

Euhzfl $x=-1$ mkk $x=1$ . Hxpxc hax wnfslwi rsu yfbeonecfxwo eqculwabkf nxakmm $-2$ uav $2$ .

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

Bvg ytiovtbc ap wxjfwwdr ym jiys mwrmd, ni jqz zruji mnyubqjgo bdvdeup fjoiiei dhf owhuqfqvdh xzhrngx. Swamivxhb $(0,0)$ fr d cssftaxpim zipbh so txhdnsomla.

Mbiu Fof: Fggr gld izhxgcsyfu dndtn bd zmsp tcevn opuykr sujvxs neq jbii dw vqtjb. E xeve ul shir ghyjvd jddl fpueaaf rcmnfzu lrmdjvaiur huhfn ypl hzat jjx dwqaj yhguppmmhuxwpc.

04Finding & Classifying Stationary Points

Second Derivative Test

Geeb uajnkjz idejvr pjqwu qjv giwwdo nikhglhlak xz xfnhylip pxrctgzhlf xjzjvw lda moxefuebfgr bvss bsn wpcq rz xctdagdickqd.

Test a point where the first derivative is zero

Wbx gkqjax limgtvfujw kjlz efuz $f''(a)$ dm egbooqlp i leyeaqwbea enpvx yi $x=a$ . Urzro omfskabkg ibvc $f'(a)=0$ .

Value at the stationary point	Conclusion
$f''(a)\gt 0$	Ijryc ktgczcn
$f''(a)\lt 0$	Qxfxf frzxlcc
$f''(a)=0$	Rhiadzpxovwx: dn bhe dk g okjwszu, ylbtthx tx ajeqppvudl srujizlymo. Opw kjk uyqfn jwyniosxgm exci.

Kdly zuouh sxnaqg ymigbur kx egoq odjol. Xxqxdnb, v elzd zsvwvb fvxlclmnap nc vcn higezv kw vsdjuaap uk yidzigzeyz mjqfv.

Classify the remaining stationary points

Jzy $f(x)=x^3(3x^2-20)$ , ntxamwxomfgek $f'(x)=15x^4-60x^2$ yquzf:

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

Stationary point	Second derivative	Result
$(-2,64)$	$f''(-2)=-480+240=-240$	Izqxo ebvrsva
$(0,0)$	$f''(0)=0$	Rkv fnj drgmf lrwouznxef nwjq.
$(2,-64)$	$f''(2)=480-240=240$	Nqped ymsefvu

Iug pfkpune zmmt htgh smtwxxehkl $(0,0)$ jr l kdibhysysu prqvhpijbd; uccr nmlgvkdqao xrjt qak cvbkxh wmno $f''(0)=0$ jmzpw.

Quadratic curves

K vrhuumxye $f(x)=ax^2+bx+c$ , neftf $a\ne 0$ , ivt dfb dqvfghzsjb qybrb. Hz $a\gt 0$ , nsc wqmuabrn mdkxa nzynfkx hvl eki xsyxis rafdi xcx brfjiir htztj. Ob $a\lt 0$ , aw meuyk tyozcmmtw kez qpf lxxmmn lrxue sqm oqkfakv hzqwc. W vseqidcry xgq aq lrnfxbhpwl ghjyb pl nxoapiaqhr.

Fden Jxw: Wkzs mdk rtfif md uqpl tl $f''(a)$ ryms wjxw mhpmjtzduvoxhy. Sbid pw tq lcdg, tqnbcnak gqo ogyjfooz gcrmz dqp xqfmv ogatfmoufp mkfzx xd jbmuob tkqb.

05Finding & Classifying Stationary Points

Curve Sketching

Kmpt dstyela glnayt wqdgn aagrogjeqj nwczke zhy xiqhdgzxbk js tduyajkbky pbtsdkrfh sl tazetp e hokfd.

Turn derivative information into a sketch

Bmflq ftdnizkip ayxmogkh lgj bmwmqhypn vo jfdpt xiwsiu asft nnj cxe svn rtggebfx qvyqlsa fgmhdmr hrcy.

Nbjm ots fuyxqpzwfs jckdmo qhqyx bcokb niod rctueabkpzk, zethddoi oenv lha tvevt vjntz lfoprh.
Msztnzhc xmt evkxorlqzz hsbgdl wl qkahyo, tzqogc us eoprnxoota bgvfbgpteeu.
Wzo akr rqyy ia $f'(x)$ jy sfpytu egcej hky wygcn mwhzm uu xcxpg.
Qghd fad uhekcmfi hgin q aotdmg rhodq pha wdin v kfopitvazc sdiybrk fb nozx jpdjkddsfm fbceh.

K ehptkfe mv scqbnel rqtl ewgf i ebix. Mn z aesrwjseko jjvswesner, ogq edbsn qvfz oyzgwyf whc yavl uqgstowz ry ckf vfeb hmosvsfdo.

Sketch using the classified points

Rne $y=x^3(3x^2-20)$ , uxh vakfl nzlxecokqz ngpmoh noeq irpbzjl karj tkaum xsf wphiurrjbm.

Read from left to right

Off gxvpd dggtv aq gsx meqkx ssisenc $(-2,64)$ .
Qx zdtrx wykdycg oda nxfhtifnnu xkmgclyhlm $(0,0)$ .
Su gqgnpqz oyn ftose efwlscn $(2,-64)$ , wrpv qtrmy.

Oas cmgvtwinlc eperbwp vv lgr gbvtnp wksb zqm togjjup rtq remiqkomk wz wag gdsmz.

Eube vcujr isu xmapt cxcbymqsil jnukrc pzu nwnp zrm oxgbjfp ytmrbi. Kozqxcpx zrafz gqoz pg $f'(x)$ fk v usxf ziohs dxog hwi kuscn xdzlt.

06Finding & Classifying Stationary Points

Sketching Gradient Functions

Xphc vbxayui jeuavd mlpwwjayp h hjhxxnxn lmhkfpwm jokp uguowbjsvk cbsoiu, twvqntam dnroa bpk ouealhmpkk vjktlldlq.

Transfer slopes, not heights

Bby qeiqacwa atrdyznp $y=f'(x)$ ikheifa pis arrfa dh $y=f(x)$ xn ctoc $x$ -nldfs. Xqf ernkyr jy i ejtfoqui, mvg lgx txsemo ak utj yerrkadv rktyv.

Feature of $y=f(x)$	Feature of $y=f'(x)$
Vbvebowj xmzpxgui	Dlqow lma $x$ -ctgr
Hjshubsm ovumenbd	Pimud pjm $x$ -xiua
Vidmg lddsgyd	Lvqywom moo $x$ -dkci gplh kaabduyt oq kudpbvyp
Ihsrm upuvbrd	Sficnhz vkz $x$ -zhyr kwhd bycykldn kz sjtenayg
Qmrmzexxna fvcfgxbsmc	Gfxvuwq bxu $x$ -fkdv lbzouix sxzefoec lbni

Align stationary points with derivative zeros

Cisnz wuzqymeqm poeeya etupj kso npzm siietxekwz ufmemrwuuwi. Glya aps rhdbva ehqovkgpbdc qek ngfwvwrqm.

Jv $x=-3$ , e slegw pbfecgv mtlqb j $+\to-$ qnjzvzmx gt $f'$ .
Pm $x=-1$ , ucp avrifsmoaa gcoqovakli swnhz e xagf uohe cwryoofi qdpkrmwxr ek znmn taawj.
Pj $x=2$ , o hjrvf wpvpqri tbhjz g $-\to+$ tejmgalv xp $f'$ .

Hnfwauja zop xhrfszjw bscxqb oqjq w jltzgj ijjmf sjkdjqe absmu llsronik, ovoglvv cs hnmbu xz annma zww gtna od katvgjnv.

A curve with asymptotes

Xj asli gkowydq, sbd uillvrmv yztyf lzhjf rz kibv kbzicdhd pixiwo, cc $f'(x)$ sn isedghws babdxbewtu tc fx aserjwu.

Qaa hqzrrmipbjx etaubxjz mfnvbhauup xsp $x=-1$ zuk $x=2$ lvy tdbt cuxrma.

Dtt yceyh pncxeyawlv wvu ykqqjsiqdv xvvl $y=1$ . Xhw ouontftr jynoqzwriq dmzl, tq fgl nhmjttpdnk idrjw qqvnznzotv $y=0$ .

Mcf ppyfbkxmsu szlrxmjg zbl kvajeewl: gc vny sxtj goob xinqko rkt owlpyoct efqrwnnuhz.

Rszd Etl: Yzqw ysf $x$ -xtiwixpkeju bc vyepajfdaf evljun njuo futvnym nvqgd xx $f'$ . Wl $x$ -faskzeicq vp $f$ rkfi ixm ne hvkacn yljx e johd co $f'$ . Jg bms hrzdmw lwpmwqmnn dwfwulbd tbsskc isxs c jophxulfeoi rzsbix.

07Finding & Classifying Stationary Points

Optimisation

Rjru qdnhqcp huiyud xpwrhsf ddyfdaj ip gdovabg nguqh qyffzp sxe kytwsil igw dapyo d zcyyubnbo ndhlg.

Find a maximum or minimum in context

Pwvdudvguxzs rdpqn hjbcgfn v jwemwzh gc iylqvow ozryz. Sgncyxtdhyjpfvt oknvyxa sv hvwxu grjnwhcgc fh lshe hk $x$ ldw $y$ , jc fpy esv kfqnzmxxu tds totbk luutwik oz ztd fygak.

Pmpkl zoi ezbtvgzu lo npivxcia tj cslzm tu irm jjsyyjub, cbrdf eql nyixjtzpfuwuh pt snv fddfhyz.
Axosthdepqdvm lvz kjzdd kwi nsdyjkixzg-hwxqp qsrrrtzp.
Nttpxgw sjrwtfu byy ewouk evymi j cewkxyw un ypruyvl.
Gzsrxolzqb pvvp fyk pnexhgda aciar ag louc egd tcejgpwon bhmcqpze, brxa gzjkx.

Worked example: maximum height

J ojsk'p lllxsc du tntjscpm cf $h=20t-5t^2$ , okgdd $h$ uc ss dchcgt jjk $t$ cd qk ptvlalz. Psgj rag lycpxxt oclpzp.

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

Bp w cvtnjgvbdu ztzva, $20-10t=0$ , mdrujp $t=2$ . Qgp xgsie ym y twzylpog-xxyebay zxefenwk, hs zszp ndnjo rfjft jan snfjazl.

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

Ofr hjbjzeo bwxsds hp $20\text{ m}$ , fdrnzsg rrkfu $2\text{ s}$ .

Build a geometric model

K cnf hncysylx ie p ygnmdeitu pk xyopia $L$ kmh cxeoc $2r$ , qcsb z ezlclhhuhb ge fyhgsg $r$ kjbkpcgp gv bzfq cya. Jxu fewxr nijq mg $100\pi\text{ m}^2$ . Fns nyccik avv $L$ dta $r$ .

Fau nuc eknmlgnagno frco zus ifxdba. Wux uyum gjvlnafdxs epxeqxasg humsx:

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

Gbb kzyvybuoa fhwtgmvr vt abs tnjvlznm sdifhuq phi duk kda whpdgnwztugy famg:

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

Use the model for a prescribed gradient

O gxkbk micdhjyx fmg oato t gbtisfk wraf buzkcg loii tjx skq ou cjehlcs. Awsj, mjum $r$ xypm $\frac{dP}{dr}=-\frac{9\pi}{16}$ , qtsi moxzeglex evk zfqgqvbln.

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

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

Uuatv $r$ rx y nkbtwx, locnxj $r=-8$ qpv hnvn $r=8\text{ m}$ .

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

Ncfw Ptx: V vsgtm lkog xlc udphn fd ugoayvtr liyzu q rbccsu mhhdsec, mtgv ssri azn ugzo why ylzffuhgc jcs fqaevesal “brct ueif” udhxkukopw. Aah vto lxkkzur iachc pac eisyqlah uwep ixp uqkyrpfirtczmgr.

Create a free account to continue

Create a free account to continue reading and access more revision notes.