MATHHX B

MATHHX B

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11.2 Monotoniforhold med differentialregning

Vi skal nu se hvordan man kan bestemme monotoniforhold ved hjælp af differentialregning. Vi lægger hårdt ud med en sætning:

  • Sætning 11.2.1
    Lad \(f\) være en differentiabel funktion defineret på et interval \(I\).

    Hvis \(f'(x)> 0\) for alle \(x\in I\), så er \(f\) voksende på \(I\).

    Hvis \(f'(x)< 0\) for alle \(x\in I\), så er \(f\) aftagende på \(I\).

    Hvis \(f'(x)= 0\) for alle \(x\in I\), så er \(f\) konstant på \(I\).

Vi vil nu argumentere for, hvorfor sætningen er rigtig. Vi vil se på første påstand (vi kan argumentere tilsvarende for de andre påstande):

Hvis \(f'(x)> 0\) for alle \(x\in I\), så er \(f\) voksende på \(I\).

Vi husker at \(f'(x)\) er tangentens hældning, så \(f'(x)> 0\) betyder at tangenten har en positiv hældning. Lad os derfor tegne en tangent med en positiv hældning i et punkt \(P\) (kun en linje, som kunne være en tangent – vi venter med funktionen).

(-tikz- diagram)

Vi forstiller os nu, hvordan funktion kan ligge. Det er klart at hvis den røde linje skal være tangent i \(P\), så må funktionen være voksende omkring punktet \(P\), hvor linjen er tangent, som vist her:

(-tikz- diagram)

Så når \(f'(x)>0\) må funktionen altså være voksende. Vi kan argumenter tilsvarende, når \(f'(x)<0\) eller \(f'(x)=0\). I sætningen nævnes også intervaller. Det er fordi at begreberne voksende, aftagende og konstant kun giver mening når vi har et interval. Det kan vi illustrere med situationen:

(-tikz- diagram)

Vi ser at der er en vandret tangent i \(P\), så \(f'(x)=0\) i \(P\), men funktionen er ikke konstant omkring \(P\). I sætningen kræves at \(f'(x)=0\) i et helt interval og ikke bare et enkelt \(x\) som på tegningen, så vi kan ikke bruge sætningen her.

Ved hjælp af sætning 11.2.1 kan vi nu beregne monotoniforhold for \(f\) ved at lave en fortegnsundersøgelse af \(f'(x)\) (læg mærke til: \(f'\), ikke \(f\)). Vi husker fra afsnittet om monotoniforhold fra mathhx bog 1, at monotoniintervallerne også indeholder ekstremumsstederne (de x-værdier hvor funktionen har maks/min), så vi får lukkede intervaller (medmindre funktionen ender i et åbent endepunkt selvfølgelig), når vi opskriver monotoniintervallerne.

  • Eksempel 11.2.1
    Vi vil undersøge monotoniforhold for funktionen \(f(x)=\frac {1}{3}x^3+x^2-8x-10\). Vi finder først \(f'(x)\). Vi får

    \[f'(x)=3\cdot \frac {1}{3}x^2+2x-8=x^2+2x-8.\]

    Nu laver vi en fortegnsundersøgelse af \(f'(x)=x^2+2x-8\). Vi starter som vi plejer med at finde nulpunkter. Differentialkvotienten \(f'(x)\) er et andengradspolynomium, så vi regner diskriminanten først:

    \[d=b^2-4ac=2^2-4\cdot 1\cdot (-8)=4+32=36\]

    Vi Indsætter nu i nulpunktsformlerne og ser at:

    \[x_1=\frac {-b+\sqrt {d}}{2a}=\frac {-2+\sqrt {36}}{2\cdot 1}=\frac {4}{2}=2\]

    og

    \[x_2=\frac {-b-\sqrt {d}}{2a}=\frac {-2-\sqrt {36}}{2\cdot 1}=\frac {-8}{2}=-4.\]

    Vi vælger nu nogle \(x\)-værdier, som omgiver nulpunkterne (-5, 0 og 3) og laver et sildeben for \(f'\):

    \(\begin {array}{ | c | c | c | c | c |c |} \hline x & -5 & -4 & 0 & 2 & 3 \\ \hline f'(x) & 7 & 0 & -8 & 0 & 7 \\ \hline \end {array}\)

    Af sildebenen kan vi se at:

    \(f'(x)\) er positiv for \(x\in ]-\infty ;-4[\cup ]2;\infty [\) \(f'(x)\) er negativ for \(x\in ]-4;2[\).

    Ved at benytte sætning 11.2.1 kan vi altså konkludere at

    \(f\) er voksende for \(x\in ]-\infty ;-4]\) og for \(x\in [2;\infty [\).
    \(f\) er aftagende for \(x\in [-4;2]\).

    Vi tegner nu grafen for at tjekke om det mon passer:

    (-tikz- diagram)

    Det passer jah.

Øvelse 11.2.1

Bestem ved beregning monotoniforholdene for følgende funktioner:

  • a) \(f(x)=x^3-6x^2-15x+29\)

  • b) \(f(x)=-5x+2\)

  • c) \(f(x)=-2x^2-4x-4\)

  • d) \(f(x)=x^4-4x^3\) (svær)

Løsning 11.2.1

  • a) Monotoniforhold for \(f\):
    \(f\) er voksende for \(x\in ]-\infty ;-1]\) og for \(x\in [5;\infty [\).
    \(f\) er aftagende for for \(x\in [-1;5].\)

  • b) Monotoniforhold for \(f\):
    \(f\) er aftagende.

  • c) Monotoniforhold for \(f\):
    \(f\) er voksende for \(x\in ]-\infty ;-1]\).
    \(f\) er aftagende for \(x\in [-1;\infty [\).

  • d) Monotoniforhold for \(f\):
    \(f\) er aftagende for \(x\in ]-\infty ;3]\)
    \(f\) er voksende for \(x\in [3;\infty [\)

Monotoniforhold for begrænsede funktioner

Til tider kan man komme ud for at der er begrænsninger på definitionsmængden, og det skal man være opmærksom på, når man bestemmer monotoniintervallerne.

  • Eksempel 11.2.2
    Lad \(f(x)=x^2-7\), hvor \(x\in ]1;4]\). Vi vil ved bergning bestemme monotoniforholdene.

    Vi finder først \(f'\):

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

    Vi finder så nulpunkter for \(f'(x)\):

    \begin{align} f'(x)&=0\\2x&=0\\x&=0. \end{align} Vi kan se at nulpunktet ligger udenfor intervallet hvor \(f\) er defineret, så nulpunket er slet ikke interessant. Vi skal altså bare undersøge fortegnet for \(f'\) i et vilkårligt punkt i definitionsmængden. Vi vælger 2:

    \[f'(2)=2\cdot 2=4\]

    Vi kan altså konkludere at \(f\) er voksende. Vi tegner for at tjekke:

    (-tikz- diagram)

    Det passer, \(f\) er voksende overalt.

Øvelse 11.2.2

Bestem ved beregning monotoniforholdene for følgende funktioner:

  • a) \(f(x)=-2x^3+3x^2+12x+5\), hvor \(x\in ]0;\infty [\)

  • b) \(f(x)=\sqrt {x}\), hvor \(x\in [1;2[\).

  • c) \(f(x)=\frac {1}{2}x^2-2x\), hvor \(x\in ]-\infty ;3[\)

Løsning 11.2.2

Bestem ved beregning monotoniforholdene for følgende funktioner:

  • a) Monotoniforhold for \(f\):
    \(f\) er voksende for \(x\in ]0;2]\)
    \(f\) er aftagende for \(x\in [2;\infty [\)

  • b) Monotoniforhold for \(f\):
    \(f\) er voksende.

  • c) Monotoniforhold for \(f\):
    \(f\) er aftagende for \(x\in ]-\infty ;2]\).
    \(f\) er voksende for \(x\in [2;3[\)