Question

find the domain and range f(x)=√x​

Answer 1

$\huge{\boxed{ \tt{Solution :)}}}$

Given ,

The function is f(x) = √x

For f(x) to be defined ,

$x \geqslant 0$

Therefore , domain = [0 , infinity )

Let , f(x) = y

Then ,

➡ y = √x

➡ x = (y)²

Since , the square of positive number or negative number is always a positive number

(y)² $\geqslant$ 0

Therefore , range = [0 , infinity )

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Answer 2

Given function is,

$\longrightarrow f(x)=\sqrt x$

We know only non - negative numbers have real square root and square root of negative numbers are not real.

Hence the domain of $f$ should be set of non - negative real numbers.

$\longrightarrow\underline{\underline{x\in[0,\ \infty)}}$

Let $x=y^2.$ Then,

$\longrightarrow f(y)=\sqrt {y^2}\quad\quad\dots(1)$

We know a positive real number should have two square roots - one is positive and the other is negative.

Both are same in magnitude, but have opposite sign.

But actually $\sqrt x$ denotes the positive square root of $x.$

That's why the square root of a square number $x^2$ is only $x$ if $x$ is non - negative or only $-x$ if $x$ is non - positive.

$\longrightarrow \sqrt{x^2}=\left\{\begin{array}{cc}x,&x\geq0\\-x,&x\leq0\end{array}\right.$

This is simply,

$\longrightarrow\sqrt{x^2}=|x|$

Then (1) becomes,

$\longrightarrow f(y)=|y|$

We know range of a modulus function is the set of non - negative real numbers.

$\longrightarrow f(y)\in[0,\ \infty)$

Or,

$\longrightarrow\underline{\underline{f(x)\in[0,\ \infty)}}$