MATHHX B
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2.3 Definitions- og værdimængde
Vi skal nu lære nogle nye begreber, I sikkert ikke har set før. Når man indføre nye begreber mere formelt, skriver man dem som en ”Definition”. En definition er altså med måde at fastlægge betydningen at et eller flere begreber.
-
Definition 2.3.1
Lad \(f\) være en funktion.
Definitionsmængden for \(f\) er mængden af mulige \(x\)’er og betegnes \(\Dm (f)\).
Værdimængden for \(f\) er mængden af mulige \(y\)’er og betegnes \(\Vm (f)\).
Lad os tage et konkret eksempel. Betragt grafen for en funktion \(f\):
Vi husker, at en ”hul” bolle betyder, at punktet ikke ligger på grafen, mens en ”udfyldt” bolle betyder, at punktet ligger på grafen, så i dette eksempel er
\[\Dm (f)=]1;7]\qquad \text {og}\qquad \Vm (f)=[2;6]\]
Læg mærke til at værdimængden starter i \(y=2\) (det laveste punkt på grafen).
Er et endepunkt på en graf ikke markeret med en bolle, betyder det, at funktionen ikke stopper, men fortsætter ud af koordinatsystemet.
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Eksempel 2.3.1
Betragt funktionen
Vi kan se, at funktionen fortsætter ned til højre, og der er derfor ingen begrænsning på definitions- og værdimængden i den retning. Så
\[\Dm (f)=[-4;\infty [ \quad \text {og} \quad \Vm (f)=]-\infty ;3]\]
Øvelse 2.3.1
Betragt graferne for to funktioner \(f\) og \(g\):
Definitionsmængde ud fra forskriften
Man kan bestemme definitionsmængden ud fra forskriften. Lad os tage nogle eksempler for at se hvordan.
-
Eksempel 2.3.2
I sidste afsnit mødte vi funktionen
\[f(x)=2x \quad , \quad -2\leq x < 1\]
Vi kan aflæse direkte ud af forskriften at \(\Dm (f)=[-2;1[\), da det svarer til begrænsningen \(-2\leq x < 1\).
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Eksempel 2.3.3
Kigger vi på funktionen
\[g(x)=2x\]
så er der ingen begrænsninger på, derfor er \(\Dm (g)=\mathbb {R}\).
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Eksempel 2.3.4
Kigger vi på funktionen
\[h(x)=\sqrt {x}\]
så kunne vi godt tro at \(\Dm (h)=\mathbb {R}\), da der ikke er skrevet nogle begrænsninger på. Men denne funktion er naturligt begrænset, fordi man ikke kan tage kvadratroden af negative tal, så \(\Dm (f)=[0;\infty [\)
Øvelse 2.3.2
Lad \(f(x)=\frac {1}{x}\).
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a) Regn \(f(4)\), hvis det kan lade sig gøre.
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b) Regn \(f(-\frac {1}{2})\), hvis det kan lade sig gøre.
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c) Regn \(f(0)\), hvis det kan lade sige gøre.
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d) Bestem definitionsmængden for \(f\) og begrund dit svar.
Løsning 2.3.2
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a) \(f(4))=\frac {1}{4}\)
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b) \(f(-\frac {1}{2}))=-2\)
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c) Det kan ikke lade sig gøre, da man ikke kan dividere med \(0\).
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d) Definitionsmængden for \(f\) er alle tal udtagen \(0\). Dette kan skrives som \(\Dm (f)=\mathbb {R}\setminus \{0\}\).
Begrundelse: Man kan dividere med alle tal undtagen nul.
Øvelse 2.3.3
Lad \(f(x)=\sqrt {x-2}\).
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a) Regn \(f(6)\), hvis det kan lade sig gøre.
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b) Regn \(f(2)\), hvis det kan lade sig gøre.
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c) Regn \(f(1)\), hvis det kan lade sig gøre.
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d) Bestem \(\Dm (f)\) og begrund dit svar.
Løsning 2.3.3
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a) \(f(6)=2\)
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b) \(f(2)=0\)
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c) Det kan ikke lade sige gøre da \(f(1)=\sqrt {-1}\), hvilket ikke kan lade sig gøre, da man ikke kan tage kvadratroden af negative tal.... hmmm jeg
håber ikke, der er en eller anden fessortype i klassen, som begynder at snakke om komplekse tal.
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d) \(\Dm (f)=[2;\infty [\), da indmaden af kvadratroden (\(x-2\)) bliver under nul, hvis \(x<2\).
Øvelse 2.3.4
Bestem definitionsmængden for funktionerne
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a) \(f(x)=x^3-x^2+x+1\)
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b) \(f(x)=\sqrt {x^2+1}\quad ,\quad x<2\)
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c) \(f(x)=\frac {x}{5}\)
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d) \(f(x)=\frac {5}{x}\)
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e) \(f(x)=\frac {2x+1}{x+37}\)
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f) \(f(x)=\sqrt {-x^2+1}\) (svær)
Løsning 2.3.4
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a) \(\Dm (f)=\mathbb {R}\)
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b) \(\Dm (f)=]-\infty ;2[\)
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c) \(\Dm (f)=\mathbb {R}\)
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d) \(\Dm (f)=\mathbb {R}\setminus \{0\}\),
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e) \(\Dm (f)=\mathbb {R}\setminus \{-37\}\)
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f) \(\Dm (f)=[-1;1]\)