Follow this example. f(x) = (x+3)/5 To find its inverse, write y=f(x) y= (x+3)/5 Interchange x and y x = (y+3)/5 solve for y in terms of x 5x=y+3 y=5x-3 The inverse of f(x) is f^-1(x) = 5x-3
Maybe; the range of the original function is given, correct? If so, then calculate the range of the inverse function by using the original functions range in the original function. Those calculated extreme values are the range of the inverse function. Suppose: f(x) = x^3, with range of -3 to +3. f(-3) = -27 f(3) = 27. Let the inverse function of f(x) = g(y); therefore g(y) = y^(1/3). The range of f(y) is -27 to 27. If true, then f(x) = f(g(y)) = f(y^(1/3)) = (y^(1/3))^3 = y g(y) = g(f(x)) = g(x^3) = (x^3)^3 = x Try by substituting the ranges into the equations, if the proofs hold, then the answer is true for the function and the range that you are testing. Sometimes, however, it can be false. Look at a transcendental function.
Y=8x-3
a function y=f(x) is said to be reflexive if x=f(y). generally of the form :- y= (ax+b)/(cx-a) where a,b,c are constants. ex:- y=(3x+5)/(4x-3)
A function that, given X, will produce Y has an inverse function that will take Y and produce X. More formally:If f(x)=y, then f-1(y)=xWhere f-1() denotes the inverse function of f()
Follow this example. f(x) = (x+3)/5 To find its inverse, write y=f(x) y= (x+3)/5 Interchange x and y x = (y+3)/5 solve for y in terms of x 5x=y+3 y=5x-3 The inverse of f(x) is f^-1(x) = 5x-3
Let F(x,y) = y - x^3 Note that (-x)^3 = -(x^3) This suggests that F(-x,-y) = -F(x,y) (-x,-y) represents the point (x,y) reflected through the origin. You could say the function F has anti-point symmetry -- each point (x,y,F) is reflected through the origin at (-x, -y, -F).
f(x) is the same thing as y= example: f(x)=2x+3 OR y=2x+3
Maybe; the range of the original function is given, correct? If so, then calculate the range of the inverse function by using the original functions range in the original function. Those calculated extreme values are the range of the inverse function. Suppose: f(x) = x^3, with range of -3 to +3. f(-3) = -27 f(3) = 27. Let the inverse function of f(x) = g(y); therefore g(y) = y^(1/3). The range of f(y) is -27 to 27. If true, then f(x) = f(g(y)) = f(y^(1/3)) = (y^(1/3))^3 = y g(y) = g(f(x)) = g(x^3) = (x^3)^3 = x Try by substituting the ranges into the equations, if the proofs hold, then the answer is true for the function and the range that you are testing. Sometimes, however, it can be false. Look at a transcendental function.
3 feet in a yard :P
3 Feet in a Yard.
3 feet in a yard
To: trantancuong21@yahoo.com PIERRE DE FERMAT's last Theorem. (x,y,z,n) belong ( N+ )^4.. n>2. (a) belong Z F is function of ( a.) F(a)=[a(a+1)/2]^2 F(0)=0 and F(-1)=0. Consider two equations F(z)=F(x)+F(y) F(z-1)=F(x-1)+F(y-1) We have a string inference F(z)=F(x)+F(y) equivalent F(z-1)=F(x-1)+F(y-1) F(z)=F(x)+F(y) infer F(z-1)=F(x-1)+F(y-1) F(z-x-1)=F(x-x-1)+F(y-x-1) infer F(z-x-2)=F(x-x-2)+F(y-x-2) we see F(z-x-1)=F(x-x-1)+F(y-x-1 ) F(z-x-1)=F(-1)+F(y-x-1 ) F(z-x-1)=0+F(y-x-1 ) give z=y and F(z-x-2)=F(x-x-2)+F(y-x-2) F(z-x-2)=F(-2)+F(y-x-2) F(z-x-2)=1+F(y-x-2) give z=/=y. So F(z-x-1)=F(x-x-1)+F(y-x-1) don't infer F(z-x-2)=F(x-x-2)+F(y-x-2) So F(z)=F(x)+F(y) don't infer F(z-1)=F(x-1)+F(y-1) So F(z)=F(x)+F(y) is not equivalent F(z-1)=F(x-1)+F(y-1) So have two cases. [F(x)+F(y)] = F(z) and F(x-1)+F(y-1)]=/=F(z-1) or vice versa So [F(x)+F(y)]-[F(x-1)+F(y-1)]=/=F(z)-F(z-1). Or F(x)-F(x-1)+F(y)-F(y-1)=/=F(z)-F(z-1). We have F(x)-F(x-1) =[x(x+1)/2]^2 - [(x-1)x/2]^2. =(x^4+2x^3+x^2/4) - (x^4-2x^3+x^2/4). =x^3. F(y)-F(y-1) =y^3. F(z)-F(z-1) =z^3. So x^3+y^3=/=z^3. n>2. .Similar. We have a string inference G(z)*F(z)=G(x)*F(x)+G(y)*F(y) equivalent G(z)*F(z-1)=G(x)*F(x-1)+G(y)*F(y-1) G(z)*F(z)=G(x)*F(x)+G(y)*F(y) infer G(z)*F(z-1)=G(x)*F(x-1)+G(y)*F(y-1) G(z)*F(z-x-1)=G(x)*F(x-x-1)+G(y-x-1)*F(y) infer G(z)*F(z-x-2)=G(x)*F(x-x-2)+G(y)*F(y-x-2) we see G(z)*F(z-x-1)=G(x)*F(x-x-1)+G(y)*F(y-x-1 ) G(z)*F(z-x-1)=G(x)*F(-1)+G(y)*F(y-x-1 ) G(z)*F(z-x-1)=0+G(y)*F(y-x-1 ) give z=y. and G(z)*F(z-x-2)=G(x)*F(x-x-2)+G(y)*F(y-x-2) G(z)*F(z-x-2)=G(x)*F(-2)+G(y)*F(y-x-2) G(z)*F(z-x-2)=G(x)+G(y)*F(y-x-2) x>0 infer G(x)>0. give z=/=y. So G(z)*F(z-x-1)=G(x)*F(x-x-1)+G(y-x-1)*F(y) don't infer G(z)*F(z-x-2)=G(x)*F(x-x-2)+G(y)*F(y-x-2) So G(z)*F(z)=G(x)*F(x)+G(y)*F(y) don't infer G(z)*F(z-1)=G(x)*F(x-1)+G(y)*F(y-1) So G(z)*F(z)=G(x)*F(x)+G(y)*F(y) is not equiivalent G(z)*F(z-1)=G(x)*F(x-1)+G(y)*F(y-1) So have two cases [G(x)*F(x)+G(y)*F(y)]=G(z)*F(z) and [ G(x)*F(x-1)+G(y)*F(y-1)]=/=G(z-1)*F(z-1) or vice versa. So [G(x)*F(x)+G(y)*F(y)] - [ G(x)*F(x-1)+G(y)*F(y-1)]=/=G(z)*[F(z)-F(z-1)]. Or G(x)*[F(x) - F(x-1)] + G(y)*[F(y)-F(y-1)]=/=G(z)*[F(z)-F(z-1).] We have x^n=G(x)*[F(x)-F(x-1) ] y^n=G(y)*[F(y)-F(y-1) ] z^n=G(z)*[F(z)-F(z-1) ] So x^n+y^n=/=z^n Happy&Peace. Trần Tấn Cường.
Последнее Пьер де Ферма теоремы. (x,y,z,n) принадлежать( N+ )^4. n>2. (a) принадлежать Z F является функцией( a.) F(a)=[a(a+1)/2]^2 F(0)=0 и F(-1)=0. Рассмотрим два уравнения F(z)=F(x)+F(y) F(z-1)=F(x-1)+F(y-1) непрерывный дедуктивного рассуждения F(z)=F(x)+F(y) эквивалент F(z-1)=F(x-1)+F(y-1) F(z)=F(x)+F(y) выводить F(z-1)=F(x-1)+F(y-1) F(z-x-1)=F(x-x-1)+F(y-x-1) выводить F(z-x-2)=F(x-x-2)+F(y-x-2) мы видим, F(z-x-1)=F(x-x-1)+F(y-x-1 ) F(z-x-1)=F(-1)+F(y-x-1 ) F(z-x-1)=0+F(y-x-1 ) давать z=y и F(z-x-2)=F(x-x-2)+F(y-x-2) F(z-x-2)=F(-2)+F(y-x-2) F(z-x-2)=1+F(y-x-2) давать z=/=y. так F(z-x-1)=F(x-x-1)+F(y-x-1) не выводить F(z-x-2)=F(x-x-2)+F(y-x-2) так F(z)=F(x)+F(y) не выводить F(z-1)=F(x-1)+F(y-1) так F(z)=F(x)+F(y) не эквивалентен F(z-1)=F(x-1)+F(y-1) Таким образом, возможны два случая. [F(x)+F(y)] = F(z) и F(x-1)+F(y-1)]=/=F(z-1) или наоборот так [F(x)+F(y)]-[F(x-1)+F(y-1)]=/=F(z)-F(z-1). или F(x)-F(x-1)+F(y)-F(y-1)=/=F(z)-F(z-1). у нас есть F(x)-F(x-1) =[x(x+1)/2]^2 - [(x-1)x/2]^2. =(x^4+2x^3+x^2/4) - (x^4-2x^3+x^2/4). =x^3. F(y)-F(y-1) =y^3. F(z)-F(z-1) =z^3. так x^3+y^3=/=z^3. n>2. аналогичный непрерывный дедуктивного рассуждения G(z)*F(z)=G(x)*F(x)+G(y)*F(y) эквивалент G(z)*F(z-1)=G(x)*F(x-1)+G(y)*F(y-1) G(z)*F(z)=G(x)*F(x)+G(y)*F(y) выводить G(z)*F(z-1)=G(x)*F(x-1)+G(y)*F(y-1) G(z)*F(z-x-1)=G(x)*F(x-x-1)+G(y-x-1)*F(y) выводить G(z)*F(z-x-2)=G(x)*F(x-x-2)+G(y)*F(y-x-2) мы видим, G(z)*F(z-x-1)=G(x)*F(x-x-1)+G(y)*F(y-x-1 ) G(z)*F(z-x-1)=G(x)*F(-1)+G(y)*F(y-x-1 ) G(z)*F(z-x-1)=0+G(y)*F(y-x-1 ) давать z=y. и G(z)*F(z-x-2)=G(x)*F(x-x-2)+G(y)*F(y-x-2) G(z)*F(z-x-2)=G(x)*F(-2)+G(y)*F(y-x-2) G(z)*F(z-x-2)=G(x)+G(y)*F(y-x-2) x>0 выводить G(x)>0. давать z=/=y. так G(z)*F(z-x-1)=G(x)*F(x-x-1)+G(y-x-1)*F(y)не выводить G(z)*F(z-x-2)=G(x)*F(x-x-2)+G(y)*F(y-x-2) так G(z)*F(z)=G(x)*F(x)+G(y)*F(y) не выводить G(z)*F(z-1)=G(x)*F(x-1)+G(y)*F(y-1) так G(z)*F(z)=G(x)*F(x)+G(y)*F(y) не эквивалентен G(z)*F(z-1)=G(x)*F(x-1)+G(y)*F(y-1) Таким образом, возможны два случая. [G(x)*F(x)+G(y)*F(y)]=G(z)*F(z) и [ G(x)*F(x-1)+G(y)*F(y-1)]=/=G(z-1)*F(z-1) или наоборот. так [G(x)*F(x)+G(y)*F(y)] - [ G(x)*F(x-1)+G(y)*F(y-1)]=/=G(z)*[F(z)-F(z-1)]. или G(x)*[F(x) - F(x-1)] + G(y)*[F(y)-F(y-1)]=/=G(z)*[F(z)-F(z-1).] у нас есть x^n=G(x)*[F(x)-F(x-1) ] y^n=G(y)*[F(y)-F(y-1) ] z^n=G(z)*[F(z)-F(z-1) ] так x^n+y^n=/=z^n Счастливые и мира. Trần Tấn Cường.
PIERRE DE FERMAT's last Theorem. (x,y,z,n) belong ( N+ )^4.. n>2. (a) belong Z F is function of ( a.) F(a)=[a(a+1)/2]^2 F(0)=0 and F(-1)=0. Consider two equations F(z)=F(x)+F(y) F(z-1)=F(x-1)+F(y-1) We have a string inference F(z)=F(x)+F(y) equivalent F(z-1)=F(x-1)+F(y-1) F(z)=F(x)+F(y) infer F(z-1)=F(x-1)+F(y-1) F(z-x-1)=F(x-x-1)+F(y-x-1) infer F(z-x-2)=F(x-x-2)+F(y-x-2) we see F(z-x-1)=F(x-x-1)+F(y-x-1 ) F(z-x-1)=F(-1)+F(y-x-1 ) F(z-x-1)=0+F(y-x-1 ) give z=y and F(z-x-2)=F(x-x-2)+F(y-x-2) F(z-x-2)=F(-2)+F(y-x-2) F(z-x-2)=1+F(y-x-2) give z=/=y. So F(z-x-1)=F(x-x-1)+F(y-x-1) don't infer F(z-x-2)=F(x-x-2)+F(y-x-2) So F(z)=F(x)+F(y) don't infer F(z-1)=F(x-1)+F(y-1) So F(z)=F(x)+F(y) is not equivalent F(z-1)=F(x-1)+F(y-1) So have two cases. [F(x)+F(y)] = F(z) and F(x-1)+F(y-1)]=/=F(z-1) or vice versa So [F(x)+F(y)]-[F(x-1)+F(y-1)]=/=F(z)-F(z-1). Or F(x)-F(x-1)+F(y)-F(y-1)=/=F(z)-F(z-1). We have F(x)-F(x-1) =[x(x+1)/2]^2 - [(x-1)x/2]^2. =(x^4+2x^3+x^2/4) - (x^4-2x^3+x^2/4). =x^3. F(y)-F(y-1) =y^3. F(z)-F(z-1) =z^3. So x^3+y^3=/=z^3. n>2. .Similar. We have a string inference G(z)*F(z)=G(x)*F(x)+G(y)*F(y) equivalent G(z)*F(z-1)=G(x)*F(x-1)+G(y)*F(y-1) G(z)*F(z)=G(x)*F(x)+G(y)*F(y) infer G(z)*F(z-1)=G(x)*F(x-1)+G(y)*F(y-1) G(z)*F(z-x-1)=G(x)*F(x-x-1)+G(y-x-1)*F(y) infer G(z)*F(z-x-2)=G(x)*F(x-x-2)+G(y)*F(y-x-2) we see G(z)*F(z-x-1)=G(x)*F(x-x-1)+G(y)*F(y-x-1 ) G(z)*F(z-x-1)=G(x)*F(-1)+G(y)*F(y-x-1 ) G(z)*F(z-x-1)=0+G(y)*F(y-x-1 ) give z=y. and G(z)*F(z-x-2)=G(x)*F(x-x-2)+G(y)*F(y-x-2) G(z)*F(z-x-2)=G(x)*F(-2)+G(y)*F(y-x-2) G(z)*F(z-x-2)=G(x)+G(y)*F(y-x-2) x>0 infer G(x)>0. give z=/=y. So G(z)*F(z-x-1)=G(x)*F(x-x-1)+G(y-x-1)*F(y) don't infer G(z)*F(z-x-2)=G(x)*F(x-x-2)+G(y)*F(y-x-2) So G(z)*F(z)=G(x)*F(x)+G(y)*F(y) don't infer G(z)*F(z-1)=G(x)*F(x-1)+G(y)*F(y-1) So G(z)*F(z)=G(x)*F(x)+G(y)*F(y) is not equiivalent G(z)*F(z-1)=G(x)*F(x-1)+G(y)*F(y-1) So have two cases [G(x)*F(x)+G(y)*F(y)]=G(z)*F(z) and [ G(x)*F(x-1)+G(y)*F(y-1)]=/=G(z-1)*F(z-1) or vice versa. So [G(x)*F(x)+G(y)*F(y)] - [ G(x)*F(x-1)+G(y)*F(y-1)]=/=G(z)*[F(z)-F(z-1)]. Or G(x)*[F(x) - F(x-1)] + G(y)*[F(y)-F(y-1)]=/=G(z)*[F(z)-F(z-1).] We have x^n=G(x)*[F(x)-F(x-1) ] y^n=G(y)*[F(y)-F(y-1) ] z^n=G(z)*[F(z)-F(z-1) ] So x^n+y^n=/=z^n Happy&Peace. Trần Tấn Cường.
Function notation is a way to name a function that is defined by an equation. For an equation in x and y, the symbol f (x) replaces y and is read as "the value of f at x" or simply as "f of x."
I like back into multidimensional space where were born to be sleeping forever thousands autumn. Pierre De Fermat's last theorem. The conditions. x,y,z,n are the integers >0 and n>2. z^n=/x^n+y^n. Assumptions z^3=x^3+y^3. Therefore z=(x^3+y^3)^1/3. I define . F(x,y)=(x^3+y^3}^1/3 - [ (x-x-1)^3+(y-x-1)^3]^1/3. Therefore [z-F(x,y)]^3={ (x^3+y^3)^1/3 - (x^3+y^3}^1/3 + [ (x-x-1)^3+(y-x-1)^3] ^1/3 }^3={ [(x-x-1)^3+(y-x-1)^3]^1/3 }^3 =(x-x-1)^3+(y-x-1)^3= (y-x-1)^3-1. Because [z-F(x,y)]^3=(y-x-1)^3-1 . Attention [(y-x-1)^3-1 ] is an integer , [(y-x-1)^3-1 ]^1/3 is an irrational number therefore [(y-x-1)^3-1 ]^2/3 is an irrational number too. Example (2^3-1) is an integer , (2^3-1)^1/3 is an irrational number and (2^3-1)^2/3 is an irrational number too. Because z-F(x,y)=[y-x-1)^3-1]^1/3. Therefore z=F(x,y)+[(y-x-1)^3-1] ^1/3. Therefore z^3=[F(x,y)]^3.+3[F(x,y)]^2*[(y-x-1)^3-1]^1/3+3F(x,y)*[(y-x-1)-1]^2/3+[(y-x-1)^3-1]. Therefore 3F(x,y)*[(y-x-1)^3-1]^2/3+[F(x,y)]^3+3[F(x,y)]^2*[(y-x-1)^3-1]^1/3+[(y-x-1)^3-1] - z^3 =0 Because. z-F(x,y)=[ (y-x-1)^3-1]^1/3. Therefore F(x,y)=z - [(y-x-1)^3-1]^1/3. Therefore 3F(x,y)*[(y-x-1)^3-1]^2/3 =3z*[(y-x-1)^3-1]^2/3 -3[(y-x-1)^3-1]. Therefore. 3z*[(y-x-1)^3-1]^2/3 - 3[(y-x-1)^3-1]+[F(x,y)]^3+3[F(x,y)]^2*[y-x-1)^3-1]^1/3+[(y-x-1)^3-1] - z^3=0.. Named [F(x,y)]^3+3[F(x,y)]^2*[(y-x-1)^3-1]^1/3+[(y-x-1)^3-1] - z^3= W. We have 3z*[(y-x-1)^3-1]^2/3 is an irrational number because z is an integer and had proved [(y-x-1)^3-1]^2/3 is an irrational number. And 3[(y-x-1)^3-1] is an integer because x,y are the integers. And 3z*[(y-x-1)^3-1]^2/3 - 3[(y-x-1)^3-1]+[F(x,y)]^3+W=0. Therefore an irrational number - an integer+W=0. Therefore W is an complex irrational number. Named 3z*[(y-x-1)^3-1]^2/3 is 3z*B And Named 3[(y-x-1)^3-1] is C . Therefore 3z*B - C +W=0. Therefore 3z*B=C-W. Because z is an integer. B is an irrational number=[(y-x-1)^3-1]^2/3 Attention (an integer)^2/3 and (an integer)^2/3 is an irrational number. W is an complex irrational number. C is an integer. Therefore. an integer*an irrational number=an complex irrational number + an integer. Unreasonable. Therefore. z^3=/x^3+y^3 Similar z^n=/x^n+y^n. ISHTAR.