MATHHX B
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11.7 Beviser til funktionsundersøgelse
Toppunktsformlen
Kan I huske I de gode gamle dage på førsteår, hvor vi lærte at man kan finde toppunktet for et andengradspolynomium ved formlen:
\[T=\left (\frac {-b}{2a};\frac {-d}{4a}\right ).\]
Vi skal nu se et bevis for denne formel. Vi skal bruge differentialregning til beviset, hvilket også er årsagen til det først kommer nu og ikke under polynomier.
-
Sætning 4.3.1 (Toppunktsformlen)
Toppunktet for et andengradspolynomium \(f(x)=ax^2+bx+c\) med diskriminant \(d\), kan bestemmes ved:
\[T=\left (\frac {-b}{2a},\frac {-d}{4a}\right ).\]
-
Bevis
Vi skal finde toppunktet T:
Men det er jo det samme som at finde ekstremum! Det vil vi nu gøre på tilsvarende måde som vi plejer.
Først finder vi \(f'(x)\). Vi ved at \(f(x)=ax^2+bx+c\), så:
\[f'(x)=2ax+b\]
Vi sætter \(f'(x)=0\):
\[2ax+b=0\]
Vi trækker \(b\) fra på begge sider:
\[2ax=-b\]
og dividerer med \(2a\):
\[x=\frac {-b}{2a}\]
Vi har hermed fundet ekstremumsstedet, som må være førstekoordinaten til toppunktet. Vi kan se det passer med formlen:
\[T=\left (\textcolor {red}{\frac {-b}{2a}};\frac {-d}{4a}\right )\]
Vi regner nu andenkoordinaten. Det gør vi også på sædvanligvis, dvs. vi indsætter førstekoordinaten i forskriften:
\[f\left (\frac {-b}{2a}\right )=a\left (\frac {-b}{2a}\right )^2+b\left (\frac {-b}{2a}\right )+c\]
Vi reducerer:
\[f\left (\frac {-b}{2a}\right )=a\frac {b^2}{4a^2}+b\frac {-b}{2a}+c\]
Vi reducerer:
\[f\left (\frac {-b}{2a}\right )=\frac {b^2}{4a}+\frac {-b^2}{2a}+c\]
Vi forlænger det midterste led med \(2\) (Vi ganger med \(2\) i tæller og nævner) og vi forlænger det sidste led med \(4a\):
\[f\left (\frac {-b}{2a}\right )=\frac {b^2}{4a}+\frac {-2b^2}{4a}+\frac {4ac}{4a}\]
Vi sætter på fælles brøkstreg:
\[f\left (\frac {-b}{2a}\right )=\frac {b^2-2b^2+4ac}{4a}\]
Vi reducerer:
\[f\left (\frac {-b}{2a}\right )=\frac {-b^2+4ac}{4a}\]
Vi genkender nu tælleren som \(-d\) (vi har jo \(d=b^2-4ac\), så derfor må \(-d=-b^2+4ac\)):
\[f\left (\frac {-b}{2a}\right )=\frac {-d}{4a}\]
Vi har dermed fundet andenkoordinaten til toppunktet. Vi kan se det passer med formlen:
\[T=\left (\frac {-b}{2a};\textcolor {red}{\frac {-d}{4a}}\right )\]
Ekstrema (svært og mest for A-niveau)
-
Sætning 11.3.1
Lad \(f\) være en differentialbel funktion som er defineret på et åbent interval \(I\).
Hvis \(f\) har et ekstremum i \(x_0\in I\), så er \(f'(x_0)=0\).
-
Bevis
Vi vil se på det tilfælde hvor vi ekstremum er et maksimum. Beviset er helt tilsvarende hvis der er tale om et minimum. Så antag at \(f\) har et maksimum i \(x_0\in I\):
At \(f\) har et maksimum i \(x_0\in I\) betyder at der er et åbent interval \(J\) omkring \(x_0\) hvor \(f(x_0)\) er større end eller lig med alle de andre funktionsværdier. Altså at \(f(x_0)\geq f(x)\) for alle \(x\in J\):
Fordi \(f\) er differentialbel i \(x_0\) ved vi at følgende grænseværdi eksisterer:
\[f'(x_0)=\lim _{\Delta x\rightarrow 0}\frac {f(x_0+\Delta x)-f(x_0)}{\Delta x}\]
Vi husker at når vi har grænseværdier, så kan \(\Delta x\) både kan være positiv og negativ. Vi vil nu undersøge grænseværdien når \(\Delta x\) er negativ og derefter når \(\Delta x\) er positiv.
Antag at \(\Delta x<0\):
Da \(f(x_0)\geq f(x)\) for alle \(x\in J\), vil \(f(x_0+\Delta x)-f(x_0)\leq 0\), når \(\Delta x\) kommer tilstrækkelig tæt på nul.
Så er vi tæt på \(x_0\), vil både tæller og nævner i brøken \(\frac {f(x_0+\Delta x)-f(x_0)}{\Delta x}\) være negative, hvilket gør brøken positiv.
Heraf kan vi konkludere at \(f'(x_0)\geq 0\) da den er givet ved grænseværdien af en brøk som altid er positiv.
Antag nu at \(\Delta x>0\):
Her gælder stadigt at \(f(x_0+\Delta x)-f(x_0)\leq 0\), når \(\Delta x\) kommer tilstrækkelig tæt på nul, men nu er \(\Delta x>0\), så brøken \(\frac {f(x_0+\Delta x)-f(x_0)}{\Delta x}\) bliver negativ.
Heraf kan vi konkludere at \(f'(x_0)\leq 0\) da den er givet ved grænseværdien af en brøk som altid er negativ.
Vi har nu vist at \(f'(x_0)\geq 0\) og \(f'(x_0)\leq 0\), hvilket må betyde at \(f'(x_0)=0\).