
How do you find the domain and range of \[y = \dfrac{1}{{x + 5}}\] ?
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.
Complete step by step answer:
Given, \[y = \dfrac{1}{{x + 5}}\]. To find the domain we need to substitute for ‘x’ then we need to find the value of ‘y’. For \[x = 1\] in \[y = \dfrac{1}{{x + 5}}\]. We have,
\[y = \dfrac{1}{{1 + 5}} \\
\Rightarrow y = \dfrac{1}{6} \\
\Rightarrow y = 0.166 \\ \]
For \[x = 2\] in \[y = \dfrac{1}{{x + 5}}\]. We have,
\[y = \dfrac{1}{{2 + 5}} \\
\Rightarrow y = \dfrac{1}{7} \\
\Rightarrow y = 0.143 \\ \]
We can see that it is well defined for all positive numbers.
For \[x = - 1\] in \[y = \dfrac{1}{{x + 5}}\]. We have,
\[y = \dfrac{1}{{ - 1 + 5}} \\
\Rightarrow y = \dfrac{1}{4} \\
\Rightarrow y = 0.25 \\ \]
For \[x = - 2\] in \[y = \dfrac{1}{{x + 5}}\]. We have,
\[y = \dfrac{1}{{ - 2 + 5}} \\
\Rightarrow y = \dfrac{1}{3} \\
\Rightarrow y = 0.33 \\ \]
We can see that \[y = \dfrac{1}{{x + 5}}\] is well defined for all real numbers that is \[x \in R\].
But for \[x = - 5\] in \[y = \dfrac{1}{{x + 5}}\]. We have,
\[y = \dfrac{1}{{ - 5 + 5}} \\
\Rightarrow y= \dfrac{1}{0} \\ \]
That is it is not defined at \[x = - 5\].
Hence the domain of \[y = \dfrac{1}{{x + 5}}\] is \[x \in R - \{ - 5\} \].
That is the domain is \[\{ x \in R:x \ne - 5\} \].
Now to find range of \[y = \dfrac{1}{{x + 5}}\], we express this in term of ‘x’ equal to
\[y = \dfrac{1}{{x + 5}}\]
\[y\left( {x + 5} \right) = 1 \\
\Rightarrow yx + 5y = 1 \\
\Rightarrow yx = 1 - 5y \\ \]
Divide by ‘y’ on both side we have,
\[ \therefore x = \dfrac{{1 - 5y}}{y}\]
We can see that ‘x’ is well defined for all real numbers but at \[y = 0\] the function ‘x’ is not defined.
Hence the range is \[y \in R - \{ 0\} \]. That is the range is \[\{ y \in R:y \ne 0\} \].Domain is \[x \in R - \{ - 5\} \].
Note: 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, \[y = \dfrac{1}{{x + 5}}\]. To find the domain we need to substitute for ‘x’ then we need to find the value of ‘y’. For \[x = 1\] in \[y = \dfrac{1}{{x + 5}}\]. We have,
\[y = \dfrac{1}{{1 + 5}} \\
\Rightarrow y = \dfrac{1}{6} \\
\Rightarrow y = 0.166 \\ \]
For \[x = 2\] in \[y = \dfrac{1}{{x + 5}}\]. We have,
\[y = \dfrac{1}{{2 + 5}} \\
\Rightarrow y = \dfrac{1}{7} \\
\Rightarrow y = 0.143 \\ \]
We can see that it is well defined for all positive numbers.
For \[x = - 1\] in \[y = \dfrac{1}{{x + 5}}\]. We have,
\[y = \dfrac{1}{{ - 1 + 5}} \\
\Rightarrow y = \dfrac{1}{4} \\
\Rightarrow y = 0.25 \\ \]
For \[x = - 2\] in \[y = \dfrac{1}{{x + 5}}\]. We have,
\[y = \dfrac{1}{{ - 2 + 5}} \\
\Rightarrow y = \dfrac{1}{3} \\
\Rightarrow y = 0.33 \\ \]
We can see that \[y = \dfrac{1}{{x + 5}}\] is well defined for all real numbers that is \[x \in R\].
But for \[x = - 5\] in \[y = \dfrac{1}{{x + 5}}\]. We have,
\[y = \dfrac{1}{{ - 5 + 5}} \\
\Rightarrow y= \dfrac{1}{0} \\ \]
That is it is not defined at \[x = - 5\].
Hence the domain of \[y = \dfrac{1}{{x + 5}}\] is \[x \in R - \{ - 5\} \].
That is the domain is \[\{ x \in R:x \ne - 5\} \].
Now to find range of \[y = \dfrac{1}{{x + 5}}\], we express this in term of ‘x’ equal to
\[y = \dfrac{1}{{x + 5}}\]
\[y\left( {x + 5} \right) = 1 \\
\Rightarrow yx + 5y = 1 \\
\Rightarrow yx = 1 - 5y \\ \]
Divide by ‘y’ on both side we have,
\[ \therefore x = \dfrac{{1 - 5y}}{y}\]
We can see that ‘x’ is well defined for all real numbers but at \[y = 0\] the function ‘x’ is not defined.
Hence the range is \[y \in R - \{ 0\} \]. That is the range is \[\{ y \in R:y \ne 0\} \].Domain is \[x \in R - \{ - 5\} \].
Note: 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.
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