Find the domain of the real valued function fx-2-x-1-x

We have, f( x ) = √(2 x) + √(1 x) Case I: 2 x ≥ 0 x ≤ 2 ..... ( 1 ) Case II: 1 + x ≥ 0 x ≥ 1 ..... ( 2 ) From equation ( ...

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Standard XII

Mathematics

Definition of Functions

Find the doma...

Question

Find the domain of the real valued function $f (x) = \sqrt{2 - x} + \sqrt{1 + x}$

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Solution

We have,
$f (x) = \sqrt{2 - x} + \sqrt{1 - x}$

Case I:- $2 - x \geq 0 x \leq 2 . . . . . (1)$

Case II:- $1 + x \geq 0 x \geq - 1 . . . . . (2)$

From equation $(1) & (2)$ , we get
$- 1 \leq x \leq 2$

Hence the domain of $f (x)$ is $[- 1, 2]$ .

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Find the domain of the real valued function f(x)=√2−x+√1+x

We have,f(x)=√2−x+√1−xCase I:- 2−x≥0x≤2.....(1)Case II:- 1+x≥0x≥−1.....(2)From equation (1)&(2), we get−1≤x≤2Hence the domain of f(x) is [−1,2].

Find the domain of the real valued function $f (x) = \sqrt{2 - x} + \sqrt{1 + x}$

We have,
$f (x) = \sqrt{2 - x} + \sqrt{1 - x}$

Case I:- $2 - x \geq 0 x \leq 2 . . . . . (1)$

Case II:- $1 + x \geq 0 x \geq - 1 . . . . . (2)$

From equation $(1) & (2)$ , we get
$- 1 \leq x \leq 2$

Hence the domain of $f (x)$ is $[- 1, 2]$ .