MATHHX B

MATHHX B

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10.3 Tangentens ligning

Vi har set hvordan man finder hældningen på en tangent til en funktion. Nu skal vi se hvordan man bestemmer en ligning for en tangent.

Tangenten er en lineær funktion med forskriften \(f(x)=ax+b\), men når vi arbejder med tangenter, snakker vi om tangentens ligning i stedet for forskrift og skriver den som \(y=ax+b\).

Tangent gennem kendt punkt
  • Eksempel 10.3.1
    Lad \(f(x)=x^2\). Vi vil gerne bestemme en ligning for tangenten igennem punktet \(P(-1,1)\):

    (-tikz- diagram)

    Ligningen for tangenten har formen \(y=ax+b\)

    Vi starter med at finde \(f'(x)\), da den kan bruges til at finde tangentens hældning \(a\).

    \[f'(x)=2x^{2-1}=2x.\]

    Vi er interesseret i hældningen når \(x=-1\), så vi regner \(f'(-1)\):

    \[f'(-1)=2\cdot (-1)=-2.\]

    Vi har altså fundet hældning \(a\) til at være\(-2\).

    Vi mangler at finde skæringen med y-aksen, \(b\). Vi har lige fundet ud af at \(a=-2\) og vi ved at når \(x=-1\) så er \(y=1\) (den går jo igennem \((-1,1)\)). Disse informationer indsætter vi i ligningen \(y=ax+b\):

    \[1=-2\cdot (-1)+b.\]

    Vi isolerer \(b\) og får \(b=-1\). Vi indsætter \(a\) og \(b\) i tangentens ligning \(y=ax+b\) og får vores facit

    \[y=-2x-1\]

    Vi tjekker at det passer med grafen. Jooo den er god nok. Tangenten har en hældning på \(-2\) og skærer \(y\)-aksen i \(-1\).

Øvelse 10.3.1

Bestem ligningen for følgende tangenter:

  • a) Tangenten igennem \((1,1)\) når \(f(x)=x^2\)

  • b) Tangenten igennem \((1,0)\) når \(f(x)=x^2-x\).

  • c) Tangenten igennem \((4,f(4))\) når \(f(x)=\sqrt {x}\).

  • d) Tangenten igennem \((2,f(2))\) når \(f(x)=\ln (x)\).

Løsning 10.3.1

  • a) \(y=2x-1\)

  • b) \(y=x-1\)

  • c) \(y=\frac {1}{4}x+1\)

  • d) \(y=\frac {1}{2}x-0,31\)

Der findes en nemmere og hurtigere metode til at bestemme tangenter. Man kan bruge følgende sætning som også står i formelsamlingen.

  • Sætning 10.3.1
    Lad \(f\) være en differentiabel funktion, og lad der være givet en tangent igennem et punkt \(P(x_0,f(x_0))\):

    (-tikz- diagram)

    Da er tangentens ligningen givet ved:

    \[y=f'(x_0)(x-x_0)+f(x_0).\]

  • Eksempel 10.3.2
    Lad \(f(x)=2x^3\). Vi vil nu bruge sætning 10.3.1 til at bestemme tangenten gennem \((2,f(2))\). Ifølge sætningen er tangenten givet ved:

    \[y=f'(x_0)(x-x_0)+f(x_0).\]

    Det første vi skal bruge er \(x_0\), som er førstekoordinaten til det punkt tangenten går igennem. I vores tilfælde er punktet \((2,f(2))\), så

    \[x_0=2\]

    Vi finder nu \(f(x_0)\):

    \[f(x_0)=f(2)=2\cdot 2^3=16\]

    Vi finder så \(f'(x)\):

    \[f'(x)=2\cdot 3x^{3-1}=6x^2.\]

    Vi kan nu finde \(f'(x_0):\)

    \[f'(x_0)=f'(2)=6\cdot 2^2=24.\]

    Vi sætter vores resultater ind i formlen: \(y=f'(x_0)(x-x_0)+f(x_0)\):

    \[y=24(x-2)+16=24x-48+16=24x-32.\]

    Altså tangentens ligning er \(y=24x-32\).

Øvelse 10.3.2

Benyt sætning 10.3.1 til at bestemme:

  • a) Tangenten gennem \((3,f(3))\) når \(f(x)=3x^2-12x+1\).

  • b) Tangenten gennem \((-1,-1)\) når \(f(x)=\frac {1}{x}\).

  • c) Tangenten gennem \((4,7)\) når \(f(x)=2x-1\).

  • d) Tangenten gennem \((9,f(9))\) når \(f(x)=6\cdot \sqrt {x}\).

Løsning 10.3.2

  • a) \(y=6x-26\)

  • b) \(y=-x-2\)

  • c) \(y=2x-1\)

  • d) \(y=x+9\)

Tangent igennem ukendt punkt, men med kendt hældning
  • Eksempel 10.3.3
    Lad \(f(x)=-x^2\). Vi vil finde den tangent som har hældningen \(1\), hvis der overhovedet findes sådan en?

    I forhold til den opgave vi liger har regnet, så kender vi pludselig ikke kender førstekoordinaten \(x_0\). Men vi kan finde \(x_0\) fordi vi ved at hældningen er \(1\). Da hældning er givet ved \(f'\), kan vi altså finde \(x_0\) ved at løse ligningen

    \[f'(x_0)=1.\]

    Vi finder \(f'\):

    \[f'(x)=-2x,\]

    så \(f'(x_0)=-2x_0,\) og ligningen vi skal løse er:

    \[-2x_0=1\]

    Ligningen har løsningen \(x_0=-\frac {1}{2}\).

    Nu hvor vi har \(x_0\) kan vi finde tangentens ligning som vi har gjort i de tidligere øvelser.

    \[f(x_0)=-\left (-\frac {1}{2}\right )^2=-\frac {1}{4}.\]

    Da hældningen på tangenten skal være \(1\) er:

    \[f'(x_0)=1\]

    Fra sætning 10.3.1 ved vi at ligningen for tangenten er givet ved:

    \[y=f'(x_0)(x-x_0)+f(x_0).\]

    Vi indsætter vores værdier og får

    \[y=1 \left (x-\left (-\frac {1}{2}\right )\right )+\left (-\frac {1}{4}\right ).\]

    Vi reducerer og får tangentens ligning til:

    \[y=x+\frac {1}{4}.\]

Øvelse 10.3.3

  • a) Lad \(f(x)=x^2+2x\). Bestem en ligning for den tangent som har en hældning på \(-4\).

  • b) Lad \(f(x)=\sqrt {x}+1\). Bestem en ligning for den tangent som er parallel med linjen \(y=\frac {1}{2}x+2\) (VINK: hvad skal hældningen på tangenten være hvis den skal være parallel med linjen?).

  • c) Lad \(f(x)=x^3\). Bestem ligningen for de to tangenter som er parallelle med linjen \(y=12x-5\).

  • d) Lad \(f(x)=5x\). Undersøg ved beregning om der findes nogle tangenter med hældning \(-3\).

Løsning 10.3.3

  • a) \(y=-4x-9\)

  • b) \(y=\frac {1}{2}x+\frac {3}{2}\)

  • c) \(y=12x+16\) og \(y=12x-16\).

  • d) Det gør der ikke.

Ekstra

Øvelse 10.3.4 (Svær)

Lad \(f(x)=x^2-2x+1\). Funktionen har to tangenter som går igennem punktet \((2,-3)\).

  • a) Beregn en ligning for de to tangenter.

VINK: Tag udgangspunkt i sætning 10.3.1.

Løsning 10.3.4

  • a) \(y=-2x+1\) og \(y=6x-15\)