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prove that the function f:R on Rdefined as f(x) = (x-1)(x-2)(x-3) is onto but not one - one.let a real valued function f be defined as f(x) = e to the power x + e to the power -x/e to the power x - e to the power -x.find it s inverse.

prove that the function f:R on Rdefined as f(x) =  (x-1)(x-2)(x-3) is onto but not one - one.let a real valued function f be defined as f(x) = e to the power x + e to the power -x/e to the power x - e to the power -x.find it s inverse.

Grade:

2 Answers

India VK
42 Points
8 years ago

For first part, notice that f(+inf) = inf and f(-inf) = -inf

Since f is continuous, therefore all points in (-inf, +inf) have a pre-image. So function is onto.

Its not one-to-one because f(1)=f(2)=f(3) = 0, so one image has multiple preimages.

 

jitender lakhanpal
62 Points
8 years ago

hi

A1) f:R ON R means domain R is mapped to codoamin R

it''s domain is R and range also is R  (it''s a polynomial function it''s domain and range both are R)

so range  = codomain that''s why this is onto function.

but not one to one

f(x) = 0 for x = 1 , 2 , 3   so for 3 values there is one value 0 .

A2)    f(x) = (e^2x+1)/(e^2x - 1)    by some algebraic manipulation

(e^2x - 1)y = (e^2x+1)

e^2x = (y+1)/(y-1)

take natural logarithms on both sides

x = 1/2 ln {(y+1)/(y-1)}

interchange x and y

y = 1/2 ln {(x+1)/(x-1)}   is inverse

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