Question

Find domain and range of real function
f(x) = x^2 + x

Please give the correct answer

Answer 1

Answer:

Step-by-step explanation:

We have,

$\tt{f(x)=x^2+x}$

Since the given function is a polynomial function, so,

$\sf{\large{\bold{\green{dom(f)\in\,\mathbb{R}}}}}$

Now, let $\sf{\blue{y=x^2+x}}$

This is in the form of $\sf{ax^2+bx+c}$, where, $\sf{a=1}$

Here, $\sf{a>0}$, so, $\sf{y=x^2+x}$ is a upward parabola

So, its range will be $\sf{\bigg[-\dfrac{D}{4a},\infty\bigg) }$

$\sf{\therefore\,D=1-4=-3}$

And,

$\sf{\dfrac{D}{4a}=\dfrac{-3}{4}}$

So,

$\sf{\large{\bold{\green{range(f)\in\,\bigg[\dfrac{3}{4} ,\infty\bigg)}}}}$

Answer 2

$\large\underline{\sf{Solution-}}$

Given function is

$\rm :\longmapsto\:f(x) = {x}^{2} + x$

Domain is defined as set of those values of x for which function is well defined.

As given function f(x) is a polynomial function.

So, Domain of f(x) is

$\bf\implies \:\boxed{ \tt{ \: x \: \in \: R \: }}$

Range of the function f(x)

Given function is

$\rm :\longmapsto\:f(x) = {x}^{2} + x$

can be rewritten as

$\rm :\longmapsto\:f(x) = {x}^{2} + x + \dfrac{1}{4} - \dfrac{1}{4}$

can further re - written as

$\rm :\longmapsto\:f(x) = {x}^{2} + 2 \times \dfrac{1}{2} \times x + \dfrac{1}{4} - \dfrac{1}{4}$

$\rm :\longmapsto\:f(x) = {\bigg[x + \dfrac{1}{2} \bigg]}^{2} - \dfrac{1}{4}$

$\bf\implies \:f(x) \geqslant \: - \: \dfrac{1}{4}$

$\bf\implies \:f(x) \: \in \: \bigg[ - \: \dfrac{1}{4}, \: \infty \bigg)$

Additional Information :-

$\boxed{ \tt{ \: |x| < y \: \implies \: - y < x < y \: \: }}$

$\boxed{ \tt{ \: |x| \leqslant y \: \implies \: - y \leqslant x \leqslant y \: \: }}$

$\boxed{ \tt{ \: |x - z| < y \: \implies \: z - y < x < z + y \: \: }}$

$\boxed{ \tt{ \: |x - z| \leqslant y \: \implies \: z - y \leqslant x \leqslant z + y \: \: }}$