MATHHX B

MATHHX B

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6.4 Fordobling og halveringskonstant

Fordoblings og halveringskonstanter er størrelser vi knytter til eksponentielle funktioner som siger noget om deres vækst.

  • Definition 6.4.1
    Lad \(f(x)\) være en voksende funktion. Fordoblingskonstanten \(T_2\) er længden på det stykke vi skal gå ud af x-aksen for at fordoble funktionsværdien:

    (-tikz- diagram)

Halveringskonstanten defineres tilsvarende:

  • Definition 6.4.2
    Lad \(f(x)\) være en aftagende funktion. Halveringskonstanten \(T_\frac {1}{2}\) er længden på det stykke vi skal gå ud af x-aksen for at halvere funktionsværdien:

    (-tikz- diagram)

Læg mærke til at definitionerne ikke siger noget om hvor vi skal starte. Hvis f.eks. \(T_2=5\) for en funktion \(f\), så vil funktionen fordobles hver gang \(x\) vokser med \(5\) - uanset hvilken \(x\)-værdi vi starter på.

Øvelse 6.4.1

Betragt graferne for de fire eksponentielle funktioner \(f\), \(g\), \(h\) og \(i\):

(-tikz- diagram)

(-tikz- diagram)

(-tikz- diagram)

(-tikz- diagram)

  • a) Funktionerne \(f\) og \(i\) har en fordoblingskonstant,mens funktionerne \(g\) og \(h\) har en halveringskonstant. Forklar hvorfor.

  • b) Bestem \(T_2\) for \(f\).

  • c) Bestem \(T_\frac {1}{2}\) for \(g\).

  • d) Bestem \(T_\frac {1}{2}\) for \(h\).

  • e) Bestem \(T_2\) for \(i\).

Løsning 6.4.1

  • a) Funktionerne \(f\) og \(i\) er voksensende men funktionerne \(g\) og \(h\) er aftagende.

  • b) \(T_2=3\)

  • c) \(T_\frac {1}{2}=2\)

  • d) \(T_\frac {1}{2}=4\)

  • e) \(T_2=1\)

Øvelse 6.4.2

Benjamin vil gerne have du enten aflæser fordoblings eller halveringskonstanten for følgende funktion: \(f(x)=10\cdot 0{,}917^x\).

  • a) Vil du finde fordoblings eller halveringskonstanten? Argumenter for dit valg.

  • b) Bestem den (tegn grafen i GeoGebra)!

Løsning 6.4.2

  • a) Da \(0{,}917<1\) er funktionen aftagende, og derfor er det kun halveringskonstanten, der kan bestemmes.

  • b) \(T_\frac {1}{2}\approx 8\).

Der findes heldigvis formler til beregning af fordoblings og halveringskonstanten. I formlerne indgår der en funktion, vi ikke har lært om endnu, nemlig funktionen \(\ln (x)\) (læses ”L N til x” eller ”L N af x”). Denne funktion kaldes den naturlige logaritme, og den skal vi se nærmere på i næste kapitel.

  • Sætning 6.4.1
    For en en eksponentiel funktion \(f(x)=ba^x\) er fordoblingskonstanten givet ved:

    \[T_2=\frac {\ln (2)}{\ln (a)}\]

  • Sætning 6.4.2
    For en en eksponentiel funktion \(f(x)=ba^x\) er halveringskonstanten givet ved:

    \[T_\frac {1}{2}=\frac {\ln \left (\frac {1}{2}\right )}{\ln (a)}\]

Vi behøver heldigvis ikke vide noget om logaritmer for at kunne bruge formlerne. Vi indsætter bare vores tal og taster det hele ind i GeoGebra/lommeregner.

Øvelse 6.4.3 (Svær)

Det ses, at fordoblingskonstanten og halveringskonstanten kun afhænger af \(a\) og altså ikke af \(b\).

  • a) Forklar, hvorfor det ikke er så mærkeligt.

Løsning 6.4.3

  • a) I en eksponentiel funktion er det alene \(a\) som bestemmer, hvor meget funktionen vokser/aftager. Da fordoblings og halveringskonstanter kun udtrykker noget om væksten er det ikke så underligt at de kun afhænger af \(a\)

  • Eksempel 6.4.1
    Lad \(f(x)=5\cdot 0{,}95^x\). Vi vil nu beregne halveringskonstanten for \(f\):

    \[T_\frac {1}{2}=\frac {\ln (\frac {1}{2})}{\ln (a)}=\frac {\ln (\frac {1}{2})}{\ln (0{,}95)}=13{,}51\]

    Altså er halveringskonstanten \(13{,}51\).

Øvelse 6.4.4

Beregn:

  • a) \(T_\frac {1}{2}\) når \(f(x)=10\cdot 0{,}87^x\)

  • b) \(T_2\) når \(f(x)=70\cdot 3^x\)

  • c) \(T_2\) når \(f(x)=30\cdot 0{,}75^x\)

Løsning 6.4.4

  • a) \(T_\frac {1}{2}=4{,}98\)

  • b) \(T_2=0{,}63\)

  • c) Man kan ikke finde fordoblingskonstanten for en aftagende funktion dit fjols.

Øvelse 6.4.5

Vinder vender nu tilbage til øvelsen med Moores lov (øvelse 6.3.2).

  • a) Bestem fordoblingskonstanten for udviklingen.

  • b) Bekræfter vores udregninger Moores lov?

Løsning 6.4.5

  • a) \(T_2=1{,}98\)

  • b) Moores lov siger:

    Antallet af transistorer pr. mikroprocessor fordobles hvert andet år.

    Det kan omformuleres til at fordoblingskonstanten er \(2\). Vi fik den til \(1{,}98\approx 2\). Så ja, Moores lov er bekræftet. Utroligt at udviklingen af computerkraft følger en så simpel lov, hva?