MATHHX B
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3.3 Stykkevis lineære funktioner
En stykkevis lineær funktion er en funktion, hvis graf er sammensat af linjestykker.
Da grafen for en stykkevis lineær funktion består af flere dele, består forskriften også af flere dele. Funktionen fra eksempel 3.3.1 har forskriften:
\[ f(x) = \begin {cases} -\frac {1}{2}x-1 & \text {for } x<2 \\ 2x-8 & \text {for } x\geq 2 \end {cases} \]
En sådan forskrift kaldes en gaffelforskrift. Vi ser, at funktionsværdierne er bestemt ved \(-\frac {1}{2}x-1\), når \(x<2\), og \(2x-8\), når \(x\geq 2\). For at vise, hvordan forskriften virker, vil vi regne
\(f(5)\). Vi observerer, at \(5\geq 2\), så vi skal bruge udtrykket \(2x-8\) til at regne funktionsværdien:
\[ f(5)=2\cdot 5-8=2 \]
Så \(f(5)=2\), hvilket også passer med grafen fra eksemplet.
Øvelse 3.3.1
Vi fortsætter med funktionen med forskriften:
\[ f(x) = \begin {cases} -\frac {1}{2}x-1 & \text {for } x<2 \\ 2x-8 & \text {for } x\geq 2 \end {cases} \]
Regn:
-
a) \(f(-2)\)
-
b) \(f(2)\)
-
c) \(f(3)\)
3.3.1
-
a) \(f(-2)=0\)
-
b) \(f(2)=-4\)
-
c) \(f(3)=-2\)
Man kan tegne grafen for en stykkevis lineær funktion ved at tegne alle linjerne i forskriften, hvorefter man visker dele af dem ud. Til sidst kan man tilføje endepunkter. Lad os tegne grafen for funktionen med forskriften:
\[ f(x) = \begin {cases} -\frac {1}{2}x-1 & \text {for } x<2 \\ 2x-8 & \text {for } x\geq 2 \end {cases} \]
Vi starter med at tegne linjerne \(y=-\frac {1}{2}x-1\) og \(y=2x-8\):
Funktionen skal skifte ved \(x=2\), så vi fjerner grafen for \(y=2x-8\) til venstre for \(x=2\) og fjerner grafen for \(y=-\frac {1}{2}x-1\) til højre for \(x=2\).
Nu mangler vi bare at tilføje endepunkter. Vi kigger på forskriften og ser, at venstre del gælder for \(x<2\), og højre del gælder for \(x\geq 2\), så der skal være et lukket endepunkt (\(\cpointtext \)) på højre del og et åbent
endepunkt (\(\opointtext \)) på venstre del:
Øvelse 3.3.2
Tegn graferne for funktionerne:
-
a) \(f(x) = \begin {cases} 2x+5 & \text {for } x\leq -1 \\ x-4 & \text {for } x > -1 \end {cases} \)
-
b) \(g(x) = \begin {cases} -x-3 & \text {for } x<-2 \\ -1 & \text {for } -2\leq x \leq 1 \\ 4x-3 & \text {for } 1
< x < 2 \\ \end {cases} \)
3.3.2
-
a)
-
b)
-
Eksempel 3.3.2
Vi vil bestemme forskriften for den stykkevise lineære funktion givet ved grafen:
Venstre del er den konstante funktion \(f(x)=2\). Den gælder, når \(x\) ligger mellem \(-4\) og \(1\), begge inklusive, da endepunkterne er markeret med \(\cpointtext \). Så vi har en gren, som hedder:
\[ f(x) = \begin {cases} 2 & \text {for } -4\leq x\leq 1 \\ \ \\ \end {cases} \]
Vi kan bestemme forskriften for højre del ved at vælge to punkter på grafen og så bruge formlerne for \(a\) og \(b\) (sætning 3.1.1). Vi aflæser to pæne punkter på højre del:
\[ (2,3)\quad \text {og}\quad (4,0) \]
Vi indsætter i formlen for \(a\):
\(\seteqnumber{0}{3.}{0}\)
\begin{align*}
a & = \frac {y_1-y_0}{x_1-x_0}\\[10pt] & = \frac {0-3}{4-2}\\[10pt] & =-1{,}5
\end{align*}
Så beregnes \(b\):
\(\seteqnumber{0}{3.}{0}\)
\begin{align*}
b & = y_0-ax_0 \\ & =3-(-1{,}5)\cdot 2\\ & =6
\end{align*}
Altså er højre del bestemt ved \(y=-1{,}5x+6\), og da den gælder for \(x>1\) (endepunktet er markeret med \(\opointtext \)), får vi:
\[ f(x) = \begin {cases} 2 & \text {for } -4\leq x\leq 1 \\ -1{,}5x+6 & \text {for } x > 1 \\ \end {cases} \]
I ovenstående eksempel brugte vi formlerne for \(a\) og \(b\) ud fra to punkter på grafen til at fastlægge forskriften. Nogle gange kan man aflæse \(a\) og \(b\) direkte ud fra grafen og dermed slippe for at bruge formlerne.
Øvelse 3.3.3
Betragt grafen for den stykkevis lineære funktion \(f\):
3.3.3
-
a) Der er to muligheder. Sådan her:
\(f(x) = \begin {cases} 1{,}5x+5 & \text {for } x\leq -2 \\ -x & \text {for } -2< x \leq 1 \\ 2 & \text {for } 1 <x< 4 \\ \end {cases}\)
eller sådan her:
\(f(x) = \begin {cases} 1{,}5x+5 & \text {for } x<-2 \\ -x & \text {for } -2\leq x \leq 1 \\ 2 & \text {for } 1 <x< 4 \\ \end {cases}\)
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b) \(f(-20)=-25\) og \(f(1)=-1\).
Øvelse 3.3.4
Betragt grafen for den stykkevis lineære funktion \(f\):
Ved aflæsning på grafen skal du bestemme:
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a) Funktionsværdien \(f(-3)\)
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b) \(\Dm (f)\) og \(\Vm (f)\)
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c) Nulpunkter for \(f\)
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d) Fortegnsvariation for \(f\)
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e) Monotoniforhold for \(f\)
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f) Ekstrema for \(f\)
3.3.4
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a) \(f(-3)=2\)
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b) \(\Dm (f)=]-\infty ; 5[\) og \(\Vm (f)=[-2;\infty [\)
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c) Der er nulpunkter i \(x=-2\) og \(x=3\)
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d) \(f\) er positiv i \(]-\infty ;-2[\) og \(]3;5[\)
\(f\) er negativ i \(]-2;3[\)
\(f\) er nul, når \(x=-2\) og \(x=3\)
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e) \(f\) er aftagende i \(]-\infty ;-1]\)
\(f\) er voksende i \([-1;5[\)
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f) \(f\) er globalt minimum i \((-1,-2)\)
Øvelse 3.3.5
Betragt grafen for funktionen \(f\).