Find the domain and the range of the real fun | vedclass

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

Question

(A) The given real function is $f(x) = \sqrt{x-1}$.For the function to be defined in the set of real numbers,the expression inside the square root mus

Vedclass · Accepted Answer

Domain: $[1, \infty)$,Range: $[0, \infty)$

Vedclass · Answer

(A) The given real function is $f(x) = \sqrt{x-1}$.For the function to be defined in the set of real numbers,the expression inside the square root must be non-negative:$x - 1 \geq 0 \Rightarrow x \geq 1$.Therefore,the domain of $f$ is the set of all real numbers greater than or equal to $1$,i.e.,$[1, \infty)$.Since $\sqrt{x-1} \geq 0$ for all $x \geq 1$,the output values of the function are always non-negative.As $x$ takes all values in $[1, \infty)$,$\sqrt{x-1}$ takes all values in $[0, \infty)$.Thus,the range of $f$ is $[0, \infty)$.

Find the domain and the range of the real function $f$ defined by $f(x) = \sqrt{x-1}$.

Similar Questions

The domain of the real-valued function $f(x) = \frac{\sqrt{\log_{10}\left(\frac{x}{x-2}\right)}}{\sqrt{[x]^2-5[x]+6}}$ is (where $[x]$ denotes the greatest integer function):

The range of the function $f(x) = x^2 + \frac{1}{x^2+1}$ is

The range of the function $f(x) = \frac{x-3}{5-x}, x \neq 5$ is

If the function $f: R - \{ 1, - 1\} \to A$ defined by $f(x) = \frac{x^2}{1 - x^2}$ is surjective,then $A$ is equal to

If the domain of the function $f(x) = \sqrt{\log_{0.6} (\left| \frac{2x-5}{x^2-4} \right|)}$ is $(-\infty, a] \cup \{b\} \cup [c, d) \cup (e, \infty)$, then the value of $a+b+c+d+e$ is ————

Vedclass Test Series

Exam Paper Generator

Online Exam Module