MATHHX B
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3.7 Beviser til lineære funktioner (V)
Vi har lært, at en sætning er et særlig vigtigt matematisk resultat. Men hvordan ved man, at en sætning er sand i første omgang? Det ved man, fordi enhver sætning har et bevis, som garanterer at sætningen er sand. I et
bevis tager man udgangspunkt i noget man allerede ved er rigtigt, og så regner man sig frem til sætnings påstand.
I dette afsnit vil vi bevise to sætninger. Den første sætning indeholder et kendt resultat som vi dog ikke tidligere har omtalt resultatet som en ”sætning”.
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Bevis
Lad \(f\) være en lineær funktion.
Vi starter med at vise, at \(f\) skærer \(y\)-aksen i punktet \((0,b)\). Skæringspunkter med y-aksen har altid førstekoordinaten \(0\), så vi kan finde \(y\)-værdien ved at sætte \(0\) ind i forskriften:
\[f(0)=a\cdot 0+b=b.\]
Så den er god nok! Funktionen skærer \(y\)-aksen i punktet i \(b\):
Vi skal nu tjekke, at funktionen vokser med \(a\), når \(x\) vokser med \(1\). Det skal gælde uanset, hvor vi starter, så vi vælger en vilkårlig \(x\)-værdi, som vi kalder \(x_0\). Lader vi denne værdi vokse med \(1\), når vi ud til
\(x_0+1\). Funktionens vækst, når \(x\) går fra \(x_0\) til \(x_0+1\), må være det grønne stykke vist her:
Vi bruger forskriften \(f(x)=ax+b\) til at regne:
\[\mathorange {f(x_0)=ax_0+b}\]
og tilsvarende regner vi: \(\seteqnumber{0}{3.}{0}\)
\begin{align*}
f(x_0+1) & = a(x_0+1)+b\\ & =ax_0+a+b
\end{align*}
Vi kan nu regne væksten (det grønne stykke):
\(\seteqnumber{0}{3.}{0}\)
\begin{align*}
\mathgreen {f(x_0+1)-f(x_0)} & = \mathred {ax_0+a+b} -(\mathorange {ax_0+b})\\ & = ax_0+a+b-ax_0-b\\ & = a
\end{align*}
Så funktionen vokser altså med \(a\), når \(x\) vokser med \(1\).
Når man først lærer om beviser, kan det være svært at se, hvor motivationen til de forskellige skridt kommer fra. Her på MATHHX skal man ikke lære at lave sine egne beviser, og derfor
er det ikke vigtigt at forstår, hvorfor man gør det ene og det andet — det vil ofte være svært at forstå. Det vigtige er, at man forstå, hvad der sker og, hvorfor det er korrekt.
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Forudsætninger
Den næste sætning vi vil bevise har både et B-niveau og et A-niveaubevis. Begge beviser kræver kendskab til faktorisering, så læs 1.6, hvis du ikke allerede har gjort det. A-niveaubeviset kræver derudover kendskab til to ligninger med to ubekendte, så regn afsnit 1.5, hvis du ikke allerede har gjort det.
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Sætning 3.2.1
Lad \(f(x)=ax+b\) være en lineær funktion og antag, at \(f\) går igennem punkterne \(P(x_0,y_0)\) og \(Q(x_1,y_1)\):
Da er \(a\) og \(b\) givet ved:
\[a=\frac {y_1-y_0}{x_1-x_0}\qquad \textrm { og }\qquad b=y_0-ax_0\]
B-niveau først:
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Bevis
Vi laver først en tegning:
Da \(f\) går igennem punkterne \(P\) og \(Q\) må (se tegning):
\[f(x_0)=y_0\quad \textrm { og }\quad f(x_1)=y_1\]
Vi bruger nu forskriften \(f(x)=ax+b\) til at regne \(f(x_0)\) og \(f(x_1)\):
\[ax_0+b=y_0\quad \textrm { og }\quad ax_1+b=y_1\]
Vi regner nu \(y_1-y_0\):
\[y_1-y_0=ax_1+b-(ax_0+b)\]
Vi ophæver parentesen:
\[y_1-y_0=ax_1+b-ax_0-b\]
og reducerer:
\[y_1-y_0=ax_1-ax_0\]
Vi sætter \(a\) ud foran parentesen (vi faktoriserer):
\[y_1-y_0=a(x_1-x_0)\]
og formlen for \(a\) fremkommer ved at dele med \(x_1-x_0\) på begge sider:
\[a=\frac {y_1-y_0}{x_1-x_0}\]
Formlen for \(b\) er nem at vise. I starten af beviset fandt vi nemlig ud af, at
\[ax_0+b=y_0\]
Så vi får \(b\) ved at trække \(ax_0\) fra på begge sider:
\[b=y_0-ax_0\]
og det var formlen for \(b\).
A-niveau beviset skal du selv når frem til i følgende øvelse.
Øvelse 3.7.1
Lav B-niveau beviset indtil du når til linjen:
\[ax_0+b=y_0\quad \textrm { og }\quad ax_1+b=y_1\]
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a) Ud fra ovenstående to ligninger kan du finde \(a\) og \(b\). Hvad hedder det afsnit, hvor metoden du skal bruge er forklaret?
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b) Bestem \(a\) og \(b\) med metoden fra afsnittet.