1 Let f R R be defined by fx 2016 4x Prove that Imf R 2Le

1) Let f : R R be defined by f(x) = 2016 4x. Prove that Imf = R

2)Let f F(R) be defined by f(x) = x²ⁿ, where n Z⁺, and S = {y R | y 0}. Prove that Imf = S.

Solution

1)

Let, y be in R

y=2016-4x

4x=2016-y

x=504-y/4

Hence, f(x)=y

So, Imf=R

2)

f(x) = x²ⁿ

LEt, y<=0

So, y=-k ,k>=0

k=x^{2n}

k=(x^2)^n

x^2=(k)^{1/n}

|x|=sqrt{(k)^{1/n}}

|x|=sqrt{(-y)^{1/n}}

Hence, imF=S

because -x^{2n}<=0