MATHHX B

MATHHX B

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1.13 Uligheder ()

En ulighed er et udsagn, som indeholder et ulighedstegn.

  • Eksempel 1.13.1
    Her er en ulighed:

    \[ 2<4 \]

    Denne ulighed er sand, fordi \(2\) er mindre end \(4\).

    Her er en anden ulighed:

    \[ 7\geq 5x-2 \]

    Denne ulighed indeholder \(x\). Den kan derfor være enten sand eller falsk afhængigt af værdien af \(x\).

Øvelse 1.13.1

Afgør, hvilke af følgende uligheder som er sande:

  • a) \(2>5\)

  • b) \(5<5\)

  • c) \(11\leq 12\)

  • d) \(11\leq 11\)

 1.13.1

  • a) Falsk

  • b) Falsk

  • c) Sand

  • d) Sand

Man løser en ulighed ved at bestemme de \(x\)’er, som gør uligheden sand. Her kan man bruge samme metode, som man bruger til at løse ligninger, bortset fra én ting: Når man ganger eller dividerer med et negativt tal, skal man vende ulighedstegnet.

  • Eksempel 1.13.2
    Vi vil løse uligheden \(2x\leq 6\). Vi dividerer med \(2\) på begge sider af ulighedstegnet, og får:

    \[ x\leq 3\]

    Vi konkluderer, at uligheden har løsningen \(x\leq 3\).

Øvelse 1.13.2

Løs ulighederne:

  • a) \(2x\geq 6\)

  • b) \(2+x<-1\)

  • c) \(x+2>14\)

 1.13.2

  • a) \(x\geq 3\)

  • b) \(x<-3\)

  • c) \(x>12\)

  • Eksempel 1.13.3
    Vi løser uligheden \(2x+4<6x+2(x-4)\):

    \begin{align*} 2x+4&<6x+2(x-4) && (\text {uligheden skrevet op})\\ 2x+4&<6x+2x-8 && (\text {parentes ganget ud})\\ 2x+4&<8x-8 && (\text {højresiden reduceret})\\ 2x+4-8x&<-8 && (\text {trukket $8x$ fra på begge sider})\\ -6x+4&<-8 && (\text {venstresiden reduceret})\\ -6x&<-8-4 && (\text {trukket $4$ fra på begge sider})\\ -6x&<-12 && (\text {højresiden reduceret})\\ x&>2 && (\text {delt med $-6$ på begge sider}) \end{align*} Læg mærke til, hvordan vi har vendt ulighedstegnet til sidst, hvor vi dividerer med \(-6\).

Øvelse 1.13.3

Løs ulighederne, og læs facit op (læs inde i dit hoved, så du ikke forstyrrer hele klassen, selvfølgelig):

  • a) \(-x\geq 7\)

  • b) \(2-x<8\)

  • c) \((x-2)\cdot 3\leq 5(x+1)\)

  • d) \(2(2+x)-(x-1)<8\)

  • e) \(0\geq 5x+10-(x+1)\)

  • f) \(-(x+3)\cdot 2>(5+1)\cdot x\)

 1.13.3

  • a) \(x\leq -7\) (læses ”\(x\) er mindre end eller lig med \(-7\)”)

  • b) \(x>-6\) (læses ”\(x\) er større end \(-6\)”)

  • c) \(x\geq -5{,}5\) (læses ”\(x\) er større end eller lig med \(-5{,}5\)”)

  • d) \(x<3\) (læses ”\(x\) er mindre end \(3\)”)

  • e) \(x\leq -2{,}25\) (læses ”\(x\) er mindre end eller lig med \(-2{,}25\)”)

  • f) \(x<-0{,}75\) (læses ”\(x\) er mindre end \(-0{,}75\)”)

Øvelse 1.13.4

Uligheder med brøker løses på tilsvarende måde, som man løser ligninger med brøker.

Løs ulighederne:

  • a) \(3\geq \frac {x}{-5}\)

  • b) \(\frac {6}{x}<2\). Forudsæt, at \(x\) er positiv.

  • c) \(\frac {6}{x}<2\). Forudsæt, at \(x\) er negativ. ()

 1.13.4

  • a) \(x\geq -15\)

  • b) \(x>3\)

  • c) \(x<0\). Måske har du fået \(x<3\), men da \(x\) skal være negativ (forudsætning), er det kun negative tal, der kan bruges.

Løsningsmængde og grundmængde

Ligesom ligninger har løsningsmængde og grundmængde, har uligheder det også. Det fungerer helt tilsvarende.

  • Eksempel 1.13.4
    Vi bestemmer løsningsmængden for uligheden \(2x+1<3\). Det ses nemt, at løsningen er:

    \[x<1\]

    De \(x\)-værdier, som opfylder denne ulighed, vil ligge i intervallet:

    \[]-\infty ;1[\]

    Løsningsmængden \(L\) er dermed:

    \[L=]-\infty ;1[\]

    Der er ingen tal, som man ikke kan sætte ind i stedet for \(x\), og derfor er grundmængden de reelle tal:

    \[ G=\mathbb {R} \]

Øvelse 1.13.5

Betragt uligheden \(4x\geq 2x+10\).

  • a) Opskriv løsningsmængden for uligheden.

  • b) Opskriv grundmængden for uligheden.

 1.13.5

  • a) \(L=[5;\infty [\)

  • b) \(G=\mathbb {R}\)

Dobbeltuligheder

En dobbeltulighed er et udsagn med to ulighedstegn. Det kunne f.eks. se således ud:

\[3<2x+1<9\]

Tænker man lidt over det, så er det klart, at sådan en dobbeltulighed bare er en kompakt måde at skrive to uligheder på, nemlig:

\[3<2x+1\qquad \text {og}\qquad 2x+1<9\]

De to uligheder kan vi løse på normal vis (gør det!), og det giver:

\[x>1\qquad \text {og}\qquad x<4\]

Så \(x\) skal altså være større end \(1\), men mindre end \(4\). Dvs. \(x\) skal ligge imellem \(1\) og \(4\):

\[1<x<4\]

og dette er så løsningen til dobbeltuligheden.

Øvelse 1.13.6

Løs ulighederne:

  • a) \(1<x-1<5\)

  • b) \(x+1\leq 2x<x+2\)

  • c) \(-1{,}5\leq -\frac {a}{400}\leq -0{,}25\)

  • d) \(-1{,}5\leq -\frac {300}{b}\leq -0{,}25\) (start med at argumentere for, at \(b>0\))

 1.13.6

  • a) \(2<x<6\)

  • b) \(1\leq x<2\)

  • c) \(100\leq a\leq 600\)

  • d) \(b>0\), fordi \(-\frac {300}{b}\) skal være negativ for at dobbeltuligheden kan blive sand. Løsningen til uligheden er: \(200\leq b \leq 1200\).

Øvelse 1.13.7

Vi vender tilbage til dobbeltuligheden fra spørgsmål d) i ovenstående øvelse.

  • a) Hvorfor er det nødvendigt at argumentere for, at \(b\) er positiv?

  • b) Opskriv grundmængden og løsningsmængden.

 1.13.7

  • a) Du får brug for at gange med \(b\) på begge sider af ulighedstegnene. Hvis \(b<0\), ville du være nødt til at vende ulighedstegnene.

  • b) \(G=\mathbb {R}\setminus \{0\}\) og \(L=[200;1200]\)