f(x) = frac{1}{x} and g(x) = frac{1}{sqrt{x}} | vedclass

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(B) Given $f(x) = \frac{1}{x}$ where $x 
eq 0$ and $g(x) = \frac{1}{\sqrt{x}}$ where $x > 0$.$1$. Find $f(g(x))$:$f(g(x)) = f\left(\frac{1}{\sqrt{x}}

Vedclass · Accepted Answer

$f(g(x))$ and $g(f(x))$ have same domain

Vedclass · Answer

(B) Given $f(x) = \frac{1}{x}$ where $x 
eq 0$ and $g(x) = \frac{1}{\sqrt{x}}$ where $x > 0$.$1$. Find $f(g(x))$:$f(g(x)) = f\left(\frac{1}{\sqrt{x}}ight) = \frac{1}{1/\sqrt{x}} = \sqrt{x}$.The domain of $f(g(x))$ is $x > 0$ because $\sqrt{x}$ is defined for $x \geq 0$ and $g(x)$ requires $x > 0$.$2$. Find $g(f(x))$:$g(f(x)) = g\left(\frac{1}{x}ight) = \frac{1}{\sqrt{1/x}} = \sqrt{x}$.The domain of $g(f(x))$ is $x > 0$ because $f(x)$ requires $x 
eq 0$ and $\sqrt{1/x}$ requires $1/x > 0$,which implies $x > 0$.Since both $f(g(x)) = \sqrt{x}$ and $g(f(x)) = \sqrt{x}$ have the same domain $(0, \infty)$,option $B$ is correct.

$f(x) = \frac{1}{x}$ and $g(x) = \frac{1}{\sqrt{x}}$,then:

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