MATHHX A

MATHHX A

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#1.#2.#3\LWRsiunitxEND {\LWRsiunitxprintdecimalsubtwo #1,,\LWRsiunitxENDTWO \ifblank {#2}{}{{\LWRsiunitxdecimal }\LWRsiunitxprintdecimalsubtwo #2,,\LWRsiunitxENDTWO }}\) \(\newcommand {\LWRsiunitxprintdecimal }[1]{\LWRsiunitxprintdecimalsub #1...\LWRsiunitxEND }\) \(\def \LWRsiunitxnumplus #1+#2+#3\LWRsiunitxEND {\ifblank {#2}{\LWRsiunitxprintdecimal {#1}}{\ifblank {#1}{\LWRsiunitxprintdecimal {#2}}{\LWRsiunitxprintdecimal {#1}\unicode {x02B}\LWRsiunitxprintdecimal {#2}}}\LWRsiunitxdistribunit }\) \(\def \LWRsiunitxnumminus #1-#2-#3\LWRsiunitxEND {\ifblank {#2}{\LWRsiunitxnumplus #1+++\LWRsiunitxEND }{\ifblank {#1}{}{\LWRsiunitxprintdecimal {#1}}\unicode {x02212}\LWRsiunitxprintdecimal {#2}\LWRsiunitxdistribunit }}\) \(\def \LWRsiunitxnumpmmacro #1\pm #2\pm #3\LWRsiunitxEND {\ifblank {#2}{\LWRsiunitxnumminus #1---\LWRsiunitxEND }{\LWRsiunitxprintdecimal {#1}\unicode {x0B1}\LWRsiunitxprintdecimal {#2}\LWRsiunitxdistribunit }}\) \(\def \LWRsiunitxnumpm #1+-#2+-#3\LWRsiunitxEND {\ifblank 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{\morecmidrules }{}\) \(\newcommand {\specialrule }[3]{\hline }\) \(\newcommand {\addlinespace }[1][]{}\) \(\def \LWRsiunitxrangephrase { \protect \mbox {to (numerical range)} }\) \(\def \LWRsiunitxdecimal {.}\)

5.3 Beviser til potensfunktioner

Vi har lært at \(b\) er andenkoordinaten til \(x=1\) og at potensfunktioner er karakteriseret ved en procent-procent vækst. Det kan formuleres mere konkret i følgende sætning:

  • Sætning 5.3.1
    Grafen for en potensfunktion \(f(x)=b\cdot x^a\) går igennem punktet \((1,b)\) og funktionen vokser med \(((1+r)^a-1)\cdot 100\%\) når \(x\) vokser med \(r\cdot 100\%\)

Vi bemærker at \(((1+r)^a-1)\cdot 100\%\) ikke afhænger af \(x\) så det er en fast procentvis vækst.

Øvelse 5.3.1

Med inspiration fra beviset til den tilsvarende sætning for eksponentielle funktioner (se afsnit 6.6 i b-bogen)

  • a) Bevis ovenstående sætning

VINK: Du får brug for potensregnereglen \((a\cdot b)^p = a^p\cdot b^p\).

Løsning 5.3.1

  • a) Vi starter med at vise at \(f\) går igennem punktet \((1,b)\). Det er nemt. Vi regner bare funktionsværdien i \(x=1\): \(f(1)=b\cdot 1^a=b\) Altså går grafen igennem punktet \((1,b)\).

    Vi vil nu vise vækstegenskaben. Vi vælger et vilkårligt \(x_0\). Lader vi \(x_0\) vokse med med \(r\cdot 100\%\), bliver \(x_0\) til \(x_0\cdot (1+r)\). Vi regner nu den procentvise vækst (som decimaltal) i \(y\)-værdierne:

    \[\frac {\text {ny} - \text {gammel}}{\text {gammel}}\]

    Vi skal finde væksten fra \(f(x_0)\) til \(f\left (x_0\cdot (1+r)\right )\), så vi skal regne

    \[\frac {f\left (x_0\cdot (1+r)\right )-f(x_0)}{f(x_0)}\]

    Vi indsætter forskriften:

    \[\frac {b\cdot \left (x_0\cdot (1+r)\right )^a -b\cdot x_0^{a}}{b\cdot x_0^{a}}\]

    og forkorter med \(b\):

    \[\frac {\left (x_0\cdot (1+r)\right )^a -x_0^{a}}{x_0^{a}}\]

    Vi bruger potensregnereglen \((a\cdot b)^p = a^p\cdot b^p\):

    \[\frac {x_0^a\cdot (1+r)^a -x_0^{a}}{x_0^{a}}\]

    og forkorter med \(x_0^a\):

    \[\left ((1+r)^a -1\right )\]

    Altså er den procentvise vækst \(\left ((1+r)^a -1\right )\cdot 100\%\), hvilket var det vi gerne ville vise.

Funktion igennem to punkter
  • Sætning 5.1.1
    Antag at grafen for en potensfunktion \(f(x)=b\cdot x^a\) går igennem punkterne \((x_0,y_0)\) og \((x_1,y_1)\). Da kan \(a\) og \(b\) bestemmes ved formlerne:

    \[a=\frac {\ln \big (\frac {y_1}{y_0}\big )}{\ln \big (\frac {x_1}{x_0}\big )}\quad \textrm { og }\quad b=\frac {y_0}{x_0^a}\]

Øvelse 5.3.2

I denne øvelse skal du igen søge inspiration i beviset for den tilsvarende sætning for eksponentielle funktioner (se afsnit 6.6 i b-bogen)

  • a) Bevis ovenstående sætning.

VINK: Du får brug for følgende regler \(\big (\frac {a}{b}\big )^p=\frac {a^p}{b^p}\) og \(\ln (a^p)=p\cdot \ln (a)\).

Løsning 5.3.2

  • a) Øv - intet facit. Vis det til din lærer og han/hun bliver imponeret. Det gør jeg i hvert fald.