MATHHX B

MATHHX B

\(\newcommand{\footnotename}{footnote}\) \(\def \LWRfootnote {1}\) \(\newcommand {\footnote }[2][\LWRfootnote ]{{}^{\mathrm {#1}}}\) \(\newcommand {\footnotemark }[1][\LWRfootnote ]{{}^{\mathrm {#1}}}\) \(\let \LWRorighspace \hspace \) \(\renewcommand {\hspace }{\ifstar \LWRorighspace \LWRorighspace }\) \(\newcommand {\TextOrMath }[2]{#2}\) \(\newcommand {\mathnormal }[1]{{#1}}\) \(\newcommand \ensuremath [1]{#1}\) \(\newcommand {\LWRframebox }[2][]{\fbox {#2}} \newcommand {\framebox }[1][]{\LWRframebox } \) \(\newcommand {\setlength }[2]{}\) \(\newcommand {\addtolength }[2]{}\) \(\newcommand {\setcounter }[2]{}\) \(\newcommand {\addtocounter }[2]{}\) \(\newcommand {\arabic }[1]{}\) \(\newcommand {\number }[1]{}\) \(\newcommand {\noalign }[1]{\text {#1}\notag \\}\) \(\newcommand {\cline }[1]{}\) \(\newcommand {\directlua }[1]{\text {(directlua)}}\) \(\newcommand {\luatexdirectlua }[1]{\text {(directlua)}}\) \(\newcommand {\protect }{}\) \(\def \LWRabsorbnumber #1 {}\) \(\def 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(numerical range)}\TextOrMath { }{\ }}\) \(\def \LWRsiunitxdecimal {.}\)

3.7 Beviser til lineære funktioner

En sætning er et særligt 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 sætningens påstand. I dette afsnit vil vi bevise to sætninger. Den første sætning udtrykker noget, vi allerede ved om lineære funktioner, men ikke tidligere har formuleret som en sætning.

  • Sætning 3.7.1
    En lineær funktion \(f(x)=ax+b\) skærer \(y\)-aksen i \(b\), og hver gang \(x\) vokser med \(1\), vokser \(f\) med \(a\).

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

    Vi starter med at vise, at \(f\) skærer \(y\)-aksen i \(b\). 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 \(b\):

    (-tikz- diagram)

    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 \(1\), kommer 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 \(\mathorange {f(x_0)}\):

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

    Vi regner også \(\mathred {f(x_0+1)}\): \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*} \mathred {f(x_0+1)}-\mathorange {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. Det forventes ikke, at du kan lave dine egne beviser, og derfor er det ikke vigtigt, at du forstår, hvorfor man gør det ene og det andet – det vil ofte være temmelig svært at forstå. Det vigtige er, at du forstår, hvad der sker, og hvorfor det er tilladt.

Den næste sætning, vi vil bevise, har både et A- og B-niveaubevis. Begge beviser kræver kendskab til faktorisering, så regn afsnit 1.6, hvis du ikke allerede har gjort det. A-niveaubeviset kræver derudover kendskab til to ligninger med to ubekendte (afsnit 1.5).

  • Sætning 3.1.1
    Lad \(f(x)=ax+b\) være en lineær funktion og antag, at \(f\) går igennem to forskellige punkter \(P(x_0,y_0)\) og \(Q(x_1,y_1)\):

    (-tikz- diagram)

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

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

Vi tager B-niveau beviset først:

  • 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 hæver parentesen:

    \[y_1-y_0=ax_1+b-ax_0-b\]

    Vi reducerer:

    \[y_1-y_0=ax_1-ax_0\]

    Vi faktoriserer med \(a\) (sætter \(a\) ud foran en parentes):

    \[y_1-y_0=a(x_1-x_0)\]

    Formlen for \(a\) fremkommer nu 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 nå frem til. I starten af beviset fandt vi nemlig ud af, at:

    \[ax_0+b=y_0\]

    Så vi får formlen for \(b\) ved at trække \(ax_0\) fra på begge sider:

    \[b=y_0-ax_0\]

A-niveaubeviset skal du selv nå frem til i følgende øvelse.

Øvelse 3.7.1

Lav B-niveaubeviset, indtil du når frem til linjen:

\[ax_0+b=y_0\quad \textrm { og }\quad ax_1+b=y_1\]

  • a) Ud fra ovenstående to ligninger kan du bestemme \(a\) og \(b\). I hvilket afsnit er det forklaret, hvordan man gør?

  • b) Bestem \(a\) og \(b\) med metoden fra afsnittet.

 3.7.1

  • a) I afsnit 1.5 om to ligninger med to ubekendte.

  • b) \(a=\frac {y_1-y_0}{x_1-x_0}\) og \(b=y_0-ax_0\) selvfølgelig.