
How do you find the domain and range of \[y = 3\sqrt {x - 2} \] ?
Answer
547.8k+ views
Hint: The domain of a function is the complete step of possible values of the independent variable. That is the domain is the set of all possible ‘x’ values which will make the function ‘work’ and will give the output of ‘y’ as a real number. The range of a function is the complete set of all possible resulting values of the dependent variable, after we have substituted the domain. We only concentrate on square root terms.
Complete step by step answer:
Given, \[y = 3\sqrt {x - 2} \]. To find where the expression is well defined we set the radicand in \[y = 3\sqrt {x - 2} \] greater than or equal to zero. That is,
\[x - 2 \geqslant 0\]
\[\Rightarrow x \geqslant 2\]
The domain is all values of ‘x’ that make the expression defined.
That is \[[2,\infty )\].
We can write this in set builder form,
The domain is \[\{ x \in R:2 \leqslant x < \infty \} \]
The range is the set of all valid \[f(x)\] values.
Since we have domain \[\{ x \in R:2 \leqslant x < \infty \} \]
If we put \[x = 2,3,4,5,...\] in \[y = 3\sqrt {x - 2} \]
We will have \[y \geqslant 0\].
That is
Put \[x = 2\] in \[y = 3\sqrt {x - 2} \] we have,
\[y = 3\sqrt {2 - 2} \\
\Rightarrow y = 3\sqrt 0 \\
\Rightarrow y = 3 \times 0 \\
\Rightarrow y = 0 \\ \]
Put \[x = 3\] in \[y = 3\sqrt {x - 2} \] we have,
\[y = 3\sqrt {3 - 2} \\
\Rightarrow y = 3\sqrt 1 \\
\Rightarrow y = 3 \\ \]
Put \[x = 4\] in \[y = 3\sqrt {x - 2} \] we have,
\[y = 3\sqrt {4 - 2} \\
\Rightarrow y= 3\sqrt 2 \\
\Rightarrow y= 3 \times 1.414 \\
\therefore y= 4.242 \\ \]
For \[x = 5\] we will have \[y \geqslant 0\] and so on. We can say that the range is \[y \geqslant 0\]. That is all non-negative real numbers.
Hence, in set builder form is \[\{ y \in R:y \geqslant 0\} \]. This is the required range and the domain is \[\{ x \in R:2 \leqslant x < \infty \} \].
Note: We know that if we have \[\sqrt a \]. Then ‘a’ is called radicand. Also in the above domain we have a closed interval, hence we can include 3 and -2. When finding the domain remember that the denominator of a fraction cannot be zero and the number under a square root sign must be positive in this section. We generally use graphs to find the domain and range. But it is a little bit difficult to draw.
Complete step by step answer:
Given, \[y = 3\sqrt {x - 2} \]. To find where the expression is well defined we set the radicand in \[y = 3\sqrt {x - 2} \] greater than or equal to zero. That is,
\[x - 2 \geqslant 0\]
\[\Rightarrow x \geqslant 2\]
The domain is all values of ‘x’ that make the expression defined.
That is \[[2,\infty )\].
We can write this in set builder form,
The domain is \[\{ x \in R:2 \leqslant x < \infty \} \]
The range is the set of all valid \[f(x)\] values.
Since we have domain \[\{ x \in R:2 \leqslant x < \infty \} \]
If we put \[x = 2,3,4,5,...\] in \[y = 3\sqrt {x - 2} \]
We will have \[y \geqslant 0\].
That is
Put \[x = 2\] in \[y = 3\sqrt {x - 2} \] we have,
\[y = 3\sqrt {2 - 2} \\
\Rightarrow y = 3\sqrt 0 \\
\Rightarrow y = 3 \times 0 \\
\Rightarrow y = 0 \\ \]
Put \[x = 3\] in \[y = 3\sqrt {x - 2} \] we have,
\[y = 3\sqrt {3 - 2} \\
\Rightarrow y = 3\sqrt 1 \\
\Rightarrow y = 3 \\ \]
Put \[x = 4\] in \[y = 3\sqrt {x - 2} \] we have,
\[y = 3\sqrt {4 - 2} \\
\Rightarrow y= 3\sqrt 2 \\
\Rightarrow y= 3 \times 1.414 \\
\therefore y= 4.242 \\ \]
For \[x = 5\] we will have \[y \geqslant 0\] and so on. We can say that the range is \[y \geqslant 0\]. That is all non-negative real numbers.
Hence, in set builder form is \[\{ y \in R:y \geqslant 0\} \]. This is the required range and the domain is \[\{ x \in R:2 \leqslant x < \infty \} \].
Note: We know that if we have \[\sqrt a \]. Then ‘a’ is called radicand. Also in the above domain we have a closed interval, hence we can include 3 and -2. When finding the domain remember that the denominator of a fraction cannot be zero and the number under a square root sign must be positive in this section. We generally use graphs to find the domain and range. But it is a little bit difficult to draw.
Recently Updated Pages
Master Class 12 English: Engaging Questions & Answers for Success

Master Class 12 Social Science: Engaging Questions & Answers for Success

Master Class 12 Maths: Engaging Questions & Answers for Success

Master Class 12 Economics: Engaging Questions & Answers for Success

Master Class 12 Physics: Engaging Questions & Answers for Success

Master Class 12 Business Studies: Engaging Questions & Answers for Success

Trending doubts
Draw a neat and well labeled diagram of TS of ovary class 12 biology CBSE

RNA and DNA are chiral molecules their chirality is class 12 chemistry CBSE

RNA and DNA are chiral molecules their chirality is class 12 chemistry CBSE

Explain sex determination in humans with the help of class 12 biology CBSE

India is a sovereign socialist secular democratic republic class 12 social science CBSE

The correct structure of ethylenediaminetetraacetic class 12 chemistry CBSE

