Define the function $f: R \rightarrow R$ by $y = f(x) = x^2, x \in R$. Complete the table given below using this definition. What are the domain and range of this function? Also,draw the graph of $f$.
$x$ $-4$ $-3$ $-2$ $-1$ $0$ $1$ $2$ $3$ $4$ $y = f(x) = x^2$
| $x$ | $-4$ | $-3$ | $-2$ | $-1$ | $0$ | $1$ | $2$ | $3$ | $4$ |
| $y = f(x) = x^2$ |
(N/A) The completed table is as follows:
Domain of $f = R$ (the set of all real numbers).
Range of $f = [0, \infty)$ (the set of all non-negative real numbers).
| $x$ | $-4$ | $-3$ | $-2$ | $-1$ | $0$ | $1$ | $2$ | $3$ | $4$ |
| $y = f(x) = x^2$ | $16$ | $9$ | $4$ | $1$ | $0$ | $1$ | $4$ | $9$ | $16$ |
Domain of $f = R$ (the set of all real numbers).
Range of $f = [0, \infty)$ (the set of all non-negative real numbers).
Explore More
Similar Questions
The domain of the function $f(x) = \frac{\sqrt{1 + x} - \sqrt{1 - x}}{x}$ is
Easy
View SolutionThe range of the real valued function $f(x) = \log_3(5 + 4x - x^2)$ is
Let $A = \{10, 11, 12, 14, 26\}$ and let $f: A \rightarrow N$ be defined such that $f(a) = \text{highest prime factor of } a$,where $a \in A$. Then the range of $f$ is:
If $x \in \mathbb{R}$,then the range of $\frac{x}{x^2-5x+9}$ is
The domain of the real valued function $f(x) = \frac{\sqrt{\log _{0.5}(x-3)}}{\sqrt{x-1}}$ is
Vedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo