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MATHHX B

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3.6 Beviser til lineære funktioner

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, selvom vi ikke tidligere har omtalt resultatet som en ”sætning”.

  • Sætning 3.6.1
    En lineær funktion \(f(x)=ax+b\) skærer \(y\)-aksen i punktet \((0,b)\), og hver gang \(x\) vokser med \(1\), vokser \(f\) med \(a\).

  • Bevis 
    Lad \(f\) være en lineær funktion.

    (-tikz- diagram)

    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 \((0,b)\):

    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:

    (-tikz- diagram)

    Vi bruger forskriften \(f(x)=ax+b\) til at regne:

    \[\textcolor {TextBlue}{f(x_0)=ax_0+b}\]

    og tilsvarende regner vi: \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):

    \begin{align*} \textcolor {TextGreen}{f(x_0+1)-f(x_0)} & = \textcolor {TextRed}{ax_0+a+b} -(\textcolor {TextBlue}{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. På HHX skal man ikke lære at lave sine egne beviser, og derfor er det ikke vigtigt, hvorfor man gør det ene og det andet — det vil ofte være svært at gøre klart. Det vigtige er, at man forstå, hvad der sker og, hvorfor det er korrekt.

Det næste bevis kræver kendskab til faktorisering, så læs 1.2, hvis du ikke allerede har gjort det.

  • 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 \(Q(x_1,y_1)\):

    (-tikz- diagram)

    Da er konstanterne \(a\) og \(b\) givet ved:

    \[a=\frac {y_1-y_0}{x_1-x_0}\qquad \textrm { og }\qquad b=y_0-ax_0\]

  • Bevis 
    Vi laver først en tegning:

    (-tikz- diagram)

    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 \(b\).