
What is the domain and range of $F\left( x \right) = \sqrt {x - 3} $?
Answer
525.3k+ views
Hint: In order to find the domain and range of the given function, first compare the radicand with zero, to get the critical points. Since it's a root function, keep in mind that the value of root should always be greater than zero. Next for range, write the function in terms of another variable, solve it and get to know about its critical points and get the range.
Complete step by step solution:
We are given with the function $F\left( x \right) = \sqrt {x - 3} $.
We know that for real numbers the domain is always $\left( { - \infty ,\infty } \right)$. But, for roots we know that the radicand should always be positive to get a real number not an imaginary number.
And, for $F\left( x \right) = \sqrt {x - 3} $, comparing the radicand with greater than equal to zero because of the above reasons, and we get:
$
x - 3 \geqslant 0 \\
x \geqslant 3 \\
$
That means for the root to be positive, the domain of the function becomes $[3,\infty )$ or $x \in [3,\infty )$.
Let, $y = F\left( x \right)$ that implies, $y = \sqrt {x - 3} $.
Squaring both the sides, along with adding $3$ on both the sides, we get:
$
y = \sqrt {x - 3} \\
{\left( y \right)^2} + 3 = {\left( {\sqrt {x - 3} } \right)^2} + 3 \\
{y^2} + 3 = x - 3 + 3 \\
{y^2} + 3 = x \\
= > x = {y^2} + 3 \\
$
From this equation that is: ${y^2} + 3 = x$, we can see that whatever we take the value of $y$, either negative or positive, it will get squared and would give positive number always, that means $x$, will always be greater than or equal to zero ($x \geqslant 0$).
That means the range of the given function $y$ or $F\left( x \right)$becomes: $[0,\infty )$ or $x \in [0,\infty )$
Therefore, the domain of the function $F\left( x \right) = \sqrt {x - 3} $ is $[3,\infty )$ and range is $[0,\infty )$.
Note:
1) For the domain or range in a root function, we could not only focus on the radicand, we need to look after the root function, to get our domain or range.
2) The interval is one side closed and one side opened because the values are greater than equal to that means the value is also included, if it was only greater than that the value was not included and it will be an open bracket.
Complete step by step solution:
We are given with the function $F\left( x \right) = \sqrt {x - 3} $.
We know that for real numbers the domain is always $\left( { - \infty ,\infty } \right)$. But, for roots we know that the radicand should always be positive to get a real number not an imaginary number.
And, for $F\left( x \right) = \sqrt {x - 3} $, comparing the radicand with greater than equal to zero because of the above reasons, and we get:
$
x - 3 \geqslant 0 \\
x \geqslant 3 \\
$
That means for the root to be positive, the domain of the function becomes $[3,\infty )$ or $x \in [3,\infty )$.
Let, $y = F\left( x \right)$ that implies, $y = \sqrt {x - 3} $.
Squaring both the sides, along with adding $3$ on both the sides, we get:
$
y = \sqrt {x - 3} \\
{\left( y \right)^2} + 3 = {\left( {\sqrt {x - 3} } \right)^2} + 3 \\
{y^2} + 3 = x - 3 + 3 \\
{y^2} + 3 = x \\
= > x = {y^2} + 3 \\
$
From this equation that is: ${y^2} + 3 = x$, we can see that whatever we take the value of $y$, either negative or positive, it will get squared and would give positive number always, that means $x$, will always be greater than or equal to zero ($x \geqslant 0$).
That means the range of the given function $y$ or $F\left( x \right)$becomes: $[0,\infty )$ or $x \in [0,\infty )$
Therefore, the domain of the function $F\left( x \right) = \sqrt {x - 3} $ is $[3,\infty )$ and range is $[0,\infty )$.
Note:
1) For the domain or range in a root function, we could not only focus on the radicand, we need to look after the root function, to get our domain or range.
2) The interval is one side closed and one side opened because the values are greater than equal to that means the value is also included, if it was only greater than that the value was not included and it will be an open bracket.
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