Consider-the-function-f-x-which-satisfying-the-functional-equation-2f-x-f-1-x-x-2-1-x-R-and-g-x-3f-x-1-The-range-of-x-g-x-1-g-x-1-is-

Question Number 25023 by Tinkutara last updated on 01/Dec/17

Consider the function f (x) which

satisfying the functional equation

2 f (x) + f (1 - x) = x^{2} + 1, \forall x \in R

and g (x) = 3 f (x) + 1 . The range of

ϕ (x) = g (x) + \frac{1}{g (x) + 1} is

Answered by prakash jain last updated on 01/Dec/17

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

substitue x b y 1 - x

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

4 f (x) + 2 f (1 - x) = 2 x^{2} + 2 (m u l t i p l y (1) b y 2)

subtract

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

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

= x^{2} + 2 x

g (x) = 3 f (x) + 1 = {(x + 1)}^{2}

\emptyset (x) = {(x + 1)}^{2} + \frac{1}{{(x + 1)}^{2} + 1}

I think you can calculate range now .

Commented by Tinkutara last updated on 02/Dec/17

I already have solved upto this . I

have doubt in finding range only .

Commented by prakash jain last updated on 02/Dec/17

for x \in R

Range of ϕ (x) = [1, \infty)