The range of the real valued function f(x) = | vedclass

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

Question

(C) Given the function $f(x) = \sqrt{9-x^2}$.For $f(x)$ to be defined,the expression inside the square root must be non-negative:$9 - x^2 \geq 0$$x^2

Vedclass · Accepted Answer

$[0, 3]$

Vedclass · Answer

(C) Given the function $f(x) = \sqrt{9-x^2}$.For $f(x)$ to be defined,the expression inside the square root must be non-negative:$9 - x^2 \geq 0$$x^2 \leq 9$$-3 \leq x \leq 3$Since $x^2$ ranges from $0$ to $9$,the expression $9 - x^2$ ranges from $9 - 9 = 0$ to $9 - 0 = 9$.Therefore,$\sqrt{9 - x^2}$ ranges from $\sqrt{0}$ to $\sqrt{9}$,which is $0$ to $3$.Thus,the range of the function is $[0, 3]$.

The range of the real valued function $f(x) = \sqrt{9-x^2}$ is

Similar Questions

$A$ real-valued function $f: A \rightarrow B$ defined by $f(x) = \frac{4-x^2}{4+x^2}$ for all $x \in A$ is a bijection. If $-4 \in A$,then $A \cap B =$

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

The domain of the real valued function $f(x) = \frac{\sqrt{\log _{0.5}(x-3)}}{\sqrt{x-1}}$ is

Find the domain of the function $f(x) = \frac{1}{\sqrt{(x + 1)(e^x - 1)(x - 4)(x + 5)(x - 6)}}$.

If $[x]$ is the greatest integer function and $2[2x - 5] - 1 = 7$,then $x$ lies in

Vedclass Test Series

Exam Paper Generator

Online Exam Module