Question

What is significant about the difference between the domain of y= square root of x and y= x²?

Answer 1

given that :

1). $y=\sqrt{x}$

domain = $[0,\infty)$ and range = $[0,\infty)$

2). $y=x^2$

domain = $(-\infty,+\infty)$ and range = $[0,\infty)$

both graph have different domain but have same range, this is because as in first graph $\sqrt{x}$ is present which is taken only the positive values of x for obtaining the real values range of graph. Thus, the domain of this graph is $[0,\infty)$ and gives values as range is $[0,\infty)$ .

But in the second graph x² is present and due to square nature it can take all real values either positive or negative as domain but its range is $[0,\infty)$, as all the negative values of the x turn to positive value due to squaring so it doesn't gives negative values.

Function is given as:

$y=\sqrt[2]{x^{-2}+4}$

$y=\sqrt[2]{\frac{1}{x^2}+4}$

for real solutions of y the term inside the square root should be as:

$\frac{1}{x^2}+4\ge0$

We can se that x has degree two so for all values of x the above term gives always positive values except x=0, as this makes it infinite value i.e. undefined, so the domain of the given equation is as:

Domain = ($-\infty,+\infty$ )-[0]

Function is given as:

$y=\sqrt[2]{x^{-2}+4}$

$y=\sqrt[2]{\frac{1}{x^2}+4}$

for real solutions of y the term inside the square root should be as:

$\frac{1}{x^2}+4\ge0$

We can se that x has degree two so for all values of x the above term gives always positive values except x=0, as this makes it infinite value i.e. undefined, so the domain of the given equation is as:

Domain = ($-\infty,+\infty$ )-[0]