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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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:

Similar Questions

Let $Q$ be the set of all rational numbers in $[0,1]$ and $f:[0,1] \rightarrow [0,1]$ be defined by $f(x) = \begin{cases} x & \text{for } x \in Q \\ 1-x & \text{for } x \notin Q \end{cases}$. Then, the set $S = \{x \in [0,1] : (f \circ f)(x) = x\}$ is equal to

Let $f(x)$ and $g(x)$ be two real polynomials of degree $2$ and $1$ respectively. If $f(g(x)) = 8x^2 - 2x$ and $g(f(x)) = 4x^2 + 6x + 1$,then the value of $f(2) + g(2)$ is

If $f: R \rightarrow R$ and $g: R \rightarrow R$ are defined by $f(x) = \begin{cases} x+2, & x>0 \\ 2-x, & x \leq 0 \end{cases}$ and $g(x) = \begin{cases} x^2-2x-2, & 1 \leq x < 2 \\ x-7, & x \geq 2 \\ x+5, & x < 1 \end{cases}$, then $\lim _{x \rightarrow 0} g(f(x))$

If $f(x) = |\cos x|$ and $g(x) = [x]$,then $gof(x)$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module