martes, 14 de abril de 2020

segon valor mitx y zeros de funcions

f(s) = 0


f(x) = d_{x}[f(c)]·( x+(-s) )


s = x+( f(x)/d_{x}[f(c)] )


[∃c][ ( c [< s or c€C ) & ( f(s)/d_{x}[f(c)] ) = s+(-s) ]


f(x) = x^{n}+(-b)
[∃c][ s = b^{(1/n)}+( b^{(1/n)}+(-1)·b^{(1/n)} ) ]


f(x) = x^{2}+(-b)
x = b^{(1/2)}
c = ( b^{(1/2)}/2 )


f(x) = x^{3}+(-b)
x = b^{(1/3)}
c = ( b^{(1/3)}/3^{(1/2)} )


f(x) = x^{n}+(-b)
x = b^{(1/n)}
c = ( b^{(1/n)}/n^{(1/(n+(-1)))} )


f(x)  = e^{x}+(-b)
[∃c][ s = ln(b)+( ln(b)+(-1)·ln(b) ) ]


f(x) = e^{x}+(-b)
x = ln(b)
c = ln( b/ln(b) )


f(x)  = e^{ln(a)·x}+(-b)
[∃c][ s = ( ln(b)/ln(a) )+( ( ln(b)/ln(a) )+(-1)·( ln(b)/ln(a) ) ) ]


f(x) = e^{ln(a)·x}+(-b)
x = ( ln(b)/ln(a) )
c = ln( b/ln(b) )


f(x)  = ln(x)+(-b)
[∃c][ s = e^{b}+( e^{b}+(-1)·e^{b} ) ]


f(x) = ln(x)+(-b)
x = e^{b}
c = ( e^{b}/b )


f(x)  = ln(ax)+(-b)
[∃c][ s = e^{(b+(-1)·ln(a))}+( e^{(b+(-1)·ln(a))}+(-1)·e^{(b+(-1)·ln(a))} ) ]


f(x) = ln(ax)+(-b)
x = e^{(b+(-1)·ln(a))}
c = ( e^{(b+(-1)·ln(a))}/b )


f(x)  = (1/x)+(-b)
[∃c][ s = (1/b)+( (1/b)+(-1)·(1/b) ) ]


f(x) = (1/x)+(-b)
x = ( 1/b )
c = ( 1/b )·i


f(x)  = (1/x^{n})+(-b)
[∃c][ s = (1/b^{(1/n)})+( (1/b^{(1/n)})+(-1)·(1/b^{(1/n)}) ) ]


f(x) = (1/x^{2})+(-b)
x = ( 1/b^{(1/2)} )
c = 2^{(1/3)}·( 1/b^{(1/2)} )·e^{(1/3)·pi·i}


f(x) = (1/x^{n})+(-b)
x = ( 1/b^{(1/n)} )
c = n^{( 1/(n+1) )}·( 1/b^{(1/n)} )·e^{( 1/(n+1) )·pi·i}

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