f(x) = x^x x E [0,1]
a)Finn maksimums og minimumsverdiene og hvor de antas.
b) Skisser grafen til f i intervallet [0,1]
Derivasjon
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mente du [tex]f(x)=x^x[/tex] ?
The square root of Chuck Norris is pain. Do not try to square Chuck Norris, the result is death.
http://www.youtube.com/watch?v=GzVSXEu0bqI - Tom Lehrer
http://www.youtube.com/watch?v=GzVSXEu0bqI - Tom Lehrer
Ja det gjorde du:p hint: logaritmisk derivasjon
The square root of Chuck Norris is pain. Do not try to square Chuck Norris, the result is death.
http://www.youtube.com/watch?v=GzVSXEu0bqI - Tom Lehrer
http://www.youtube.com/watch?v=GzVSXEu0bqI - Tom Lehrer
[tex]y=x^x[/tex]
[tex]\ln(y)=x\ln(x)[/tex]
Deriverer implisitt mhp x
[tex]\frac1{y}\cdot \frac{dy}{dx}=1\cdot \ln(x)+x\cdot \frac1{x}[/tex]
[tex]\frac{dy}{dx}=(\ln(x)+1)\cdot y[/tex]
[tex]y^\prime = (\ln(x)+1)\cdot x^x[/tex]
Finner evt. maks./min. punkter ved å sette den deriverte lik null:
[tex](\ln(x)+1)\cdot x^x=0\ \Rightarrow \ x^x=0 \ \vee \ \ln(x)+1=0[/tex]
[tex]x^x\neq0[/tex]
[tex]\ln(x)=-1[/tex]
[tex]e^{\ln(x)}=e^{-1}[/tex]
[tex]x=\frac1{e^1}[/tex]
[tex]y(\frac1{e^1})={\frac1{e^1}}^{\frac1{e^1}}\approx 0.69[/tex]
Bunnpunkt: [tex](\frac1{e^1},\,0.69)[/tex]
[tex]\ln(y)=x\ln(x)[/tex]
Deriverer implisitt mhp x
[tex]\frac1{y}\cdot \frac{dy}{dx}=1\cdot \ln(x)+x\cdot \frac1{x}[/tex]
[tex]\frac{dy}{dx}=(\ln(x)+1)\cdot y[/tex]
[tex]y^\prime = (\ln(x)+1)\cdot x^x[/tex]
Finner evt. maks./min. punkter ved å sette den deriverte lik null:
[tex](\ln(x)+1)\cdot x^x=0\ \Rightarrow \ x^x=0 \ \vee \ \ln(x)+1=0[/tex]
[tex]x^x\neq0[/tex]
[tex]\ln(x)=-1[/tex]
[tex]e^{\ln(x)}=e^{-1}[/tex]
[tex]x=\frac1{e^1}[/tex]
[tex]y(\frac1{e^1})={\frac1{e^1}}^{\frac1{e^1}}\approx 0.69[/tex]
Bunnpunkt: [tex](\frac1{e^1},\,0.69)[/tex]
The square root of Chuck Norris is pain. Do not try to square Chuck Norris, the result is death.
http://www.youtube.com/watch?v=GzVSXEu0bqI - Tom Lehrer
http://www.youtube.com/watch?v=GzVSXEu0bqI - Tom Lehrer