# If $$f(x) = \sqrt{x}$$ and $$g(x) = \sqrt{3 – x}$$ then find f0g and its domain and range.

Solution:

f(x) = √x

$$g(x) = \sqrt{3-x}$$

fog(x) = ?

We know,

fog = f[g(x)] = f($$\sqrt{3-x}$$

$$= \sqrt{\sqrt{3-x}}$$

$$= (3-x)^{\frac{1}{4}}$$

Domain:

To find domain of fog,

First we have to find domain of $$g(x) = \sqrt{3-x}$$

Domain of g(x) is

3 – x ≥ 0

i.e. x ≤ 3  i.e. A = (-∞, 3].

and domain of $$f(g(x)) = (3-x)^{\frac{1}{4}}$$

3 – x ≥ 0

i.e. x ≤ 3  i.e. B = (-∞, 3].

Here, A∩B = (-∞, 3] is domain of fog.

Range:

$$y = (3-x)^{\frac{1}{4}}$$

4√y = 3 – x

x  = 3 – 4√y

x exists for y > 0 but domain limit is (-∞, 3]

Range will be (0, 3]