If the functions are defined as f ( x ) = x and g ( x ) = 1 - x , then what is the common domain of the following functions: f + g , f - g , f g , g f , g - f , where ( f g ) ( x ) = f ( x ) g ( x ) , ( f g ) ( x ) = f ( x ) g ( x ) .

Question

If the functions are defined as $f (x) = \sqrt{x}$ and $g (x) = \sqrt{1 - x}$ , then what is the common domain of the following functions: $f + g,$ $f - g,$ $\frac{f}{g},$ $\frac{g}{f},$ $g - f,$ where $(f \pm g) (x) = f (x) \pm g (x),$ $(\frac{f}{g}) (x) = \frac{f (x)}{g (x)} .$

Smart Achievers · Accepted Answer

(4)

Clearly, $f (x)$ is defined for all $x \geq 0$ and $g (x)$ is defined for all $x \leq 1$

In the common domain of $f + g, f - g, \frac{f}{g}, \frac{g}{f}, g - f,$ $f (x) \neq 0$ and $g (x) \neq 0$ $\Rightarrow x \neq 0, 1$

Hence, $x \in (0, 1)$

Solution

Q.
If the functions are defined as $f (x) = \sqrt{x}$ and $g (x) = \sqrt{1 - x}$ , then what is the common domain of the following functions: $f + g,$ $f - g,$ $\frac{f}{g},$ $\frac{g}{f},$ $g - f,$ where $(f \pm g) (x) = f (x) \pm g (x),$ $(\frac{f}{g}) (x) = \frac{f (x)}{g (x)} .$

1 $0 \leq x < 1$

2 $0 < x \leq 1$

3 $0 \leq x \leq 1$

4 $0 < x < 1$

Ans.
(4)

Clearly, $f (x)$ is defined for all $x \geq 0$ and $g (x)$ is defined for all $x \leq 1$

In the common domain of $f + g, f - g, \frac{f}{g}, \frac{g}{f}, g - f,$ $f (x) \neq 0$ and $g (x) \neq 0$ $\Rightarrow x \neq 0, 1$

Hence, $x \in (0, 1)$

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Solution

Q. If the functions are defined as f(x)=x and g(x)=1-x, then what is the common domain of the following functions: f+g,f-g, fg, gf, g-f, where (f±g)(x)=f(x)±g(x), (fg)(x)=f(x)g(x).

1 0≤x<1

2 0<x≤1

3 0≤x≤1

4 0<x<1

Ans. (4) Clearly, f(x) is defined for all x≥0 and g(x) is defined for all x≤1 In the common domain of f+g, f-g, fg, gf, g-f, f(x)≠0 and g(x)≠0⇒ x≠0,1 Hence, x∈(0,1)