
How do you find the domain and range of the inverse functions $ f(x) = \dfrac{1}{{x - 2}} $ ?
Answer
563.4k+ views
Hint: Domain refers to all the possible values that x can take, that is the values of the x for which a function is defined is called the domain of the function. On putting different values from the domain, we obtain different values of the function, thus the set of all the possible values that a function can attain is called its range. Using the above-mentioned definition of domain and range of a function, we can find out the domain and range of the given function.
Complete step-by-step answer:
We are given that $ f(x) = \dfrac{1}{{x - 2}} $
We know that for a fraction to exist, its denominator can take any value except zero, so –
$
x - 2 \ne 0 \\
\Rightarrow x \ne 2 \\
$
Let
$
f(x) = y \\
\Rightarrow y = \dfrac{1}{{x - 2}} \\
\Rightarrow x - 2 = \dfrac{1}{y} \\
\Rightarrow x = \dfrac{1}{y} + 2 \\
\Rightarrow x = \dfrac{{1 + 2y}}{y} \;
$
Now, the denominator cannot be zero, so $ y \ne 0 $
So, the domain of $ f(x) $ is all real numbers except 2 and the range of the function is all real numbers except 0.
Now, when the function $ f(x) $ is inverted, the domain of $ f(x) $ becomes the range of $ {f^{ - 1}}(x) $ and the range of $ f(x) $ becomes the domain of $ {f^{ - 1}}(x) $ .
Hence, the domain of inverse $ f(x) $ is all real numbers except 0 and the range of inverse $ f(x) $ is all the real numbers except 2.
Note: For solving this kind of questions, we must know the concept of the domain and range of function clearly. In this question, we knew that the denominator cannot be zero that’s why we could get the correct answer easily. Instead of finding the domain and range of function $ f(x) $ and then finding the domain and range of inverse $ f(x) $ , we can first find the inverse $ f(x) $ and then find its domain and range following a similar approach. In the given question, $ {f^{ - 1}}(x) = \dfrac{{1 + 2x}}{x} $ is the inverse of the function $ f(x) $ .
Complete step-by-step answer:
We are given that $ f(x) = \dfrac{1}{{x - 2}} $
We know that for a fraction to exist, its denominator can take any value except zero, so –
$
x - 2 \ne 0 \\
\Rightarrow x \ne 2 \\
$
Let
$
f(x) = y \\
\Rightarrow y = \dfrac{1}{{x - 2}} \\
\Rightarrow x - 2 = \dfrac{1}{y} \\
\Rightarrow x = \dfrac{1}{y} + 2 \\
\Rightarrow x = \dfrac{{1 + 2y}}{y} \;
$
Now, the denominator cannot be zero, so $ y \ne 0 $
So, the domain of $ f(x) $ is all real numbers except 2 and the range of the function is all real numbers except 0.
Now, when the function $ f(x) $ is inverted, the domain of $ f(x) $ becomes the range of $ {f^{ - 1}}(x) $ and the range of $ f(x) $ becomes the domain of $ {f^{ - 1}}(x) $ .
Hence, the domain of inverse $ f(x) $ is all real numbers except 0 and the range of inverse $ f(x) $ is all the real numbers except 2.
Note: For solving this kind of questions, we must know the concept of the domain and range of function clearly. In this question, we knew that the denominator cannot be zero that’s why we could get the correct answer easily. Instead of finding the domain and range of function $ f(x) $ and then finding the domain and range of inverse $ f(x) $ , we can first find the inverse $ f(x) $ and then find its domain and range following a similar approach. In the given question, $ {f^{ - 1}}(x) = \dfrac{{1 + 2x}}{x} $ is the inverse of the function $ f(x) $ .
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