
How do you find the domain of $ \left[ {fog} \right]\left( x \right) $ given $ f\left( x \right) = \dfrac{1}{x} $ and $ g\left( x \right) = 7 - x $ ?
Answer
530.4k+ views
Hint: This question deals with the concept of domain. Before solving the question, we should know what is domain. The domain refers to the set of possible values of x for which the function will be defined. Here in this question, we need to find the domain of the given function. We can write $ \left( {fog} \right)\left( x \right) $ as $ f\left( {g\left( x \right)} \right) $ . Here we have a composition of two functions. The composition is an operation where two functions say ‘ $ f $ ’ and ‘ $ g $ ’ generate a new function say ‘ $ h $ ’ in such a way that $ h\left( x \right) = g\left( {f\left( x \right)} \right) $ . It means here the function $ g $ is applied to the function of ‘ $ x $ ’.
Complete step by step solution:
As we know that, $ \left( {fog} \right)\left( x \right) = f\left( {g\left( x \right)} \right) $
Given, $ f\left( x \right) = \dfrac{1}{x} $ and $ g\left( x \right) = 7 - x $
Now,
$ \Rightarrow \left( {fog} \right)\left( x \right) = f\left( {g\left( x \right)} \right) $
We have $ f\left( x \right) = \dfrac{1}{x} $ in this we need to put $ x = g\left( x \right) $ . Then we have,
$ \Rightarrow \left( {fog} \right)\left( x \right) = \dfrac{1}{{g\left( x \right)}} $ , which is defined for all values of $ g\left( x \right) \ne 0 $
But we are given that $ g\left( x \right) = 7 - x $ . Thus, we have
$ \Rightarrow \left( {fog} \right)\left( x \right) = \dfrac{1}{{g\left( x \right)}} = \dfrac{1}{{7 - x}} $
Since,
$
\Rightarrow g\left( x \right) = 7 - x \\
\Rightarrow g\left( x \right) \ne 0 \to x \ne 7 \\
$
$ \left( {fog} \right)\left( x \right) $ is defined for all real values of $ x $ except $ x = 7 $ .
Therefore, the domain of $ \left[ {fog} \right]\left( x \right) $ is $ \mathbb{R} - \left\{ 7 \right\} $ .
Note: The given question is an easy one. Students should solve such types of questions very carefully. In order to solve similar types of questions, students should follow the same procedure for any composition problem. If ‘ $ f $ ’ and ‘ $ g $ ’ are one-one functions then the composition function $ \left( {fog} \right)\left( x \right) $ is also one-one function. The function composition of two onto function is always onto. Students can find similar types of questions in their NCERT textbooks and practise them for more clarity.
Complete step by step solution:
As we know that, $ \left( {fog} \right)\left( x \right) = f\left( {g\left( x \right)} \right) $
Given, $ f\left( x \right) = \dfrac{1}{x} $ and $ g\left( x \right) = 7 - x $
Now,
$ \Rightarrow \left( {fog} \right)\left( x \right) = f\left( {g\left( x \right)} \right) $
We have $ f\left( x \right) = \dfrac{1}{x} $ in this we need to put $ x = g\left( x \right) $ . Then we have,
$ \Rightarrow \left( {fog} \right)\left( x \right) = \dfrac{1}{{g\left( x \right)}} $ , which is defined for all values of $ g\left( x \right) \ne 0 $
But we are given that $ g\left( x \right) = 7 - x $ . Thus, we have
$ \Rightarrow \left( {fog} \right)\left( x \right) = \dfrac{1}{{g\left( x \right)}} = \dfrac{1}{{7 - x}} $
Since,
$
\Rightarrow g\left( x \right) = 7 - x \\
\Rightarrow g\left( x \right) \ne 0 \to x \ne 7 \\
$
$ \left( {fog} \right)\left( x \right) $ is defined for all real values of $ x $ except $ x = 7 $ .
Therefore, the domain of $ \left[ {fog} \right]\left( x \right) $ is $ \mathbb{R} - \left\{ 7 \right\} $ .
Note: The given question is an easy one. Students should solve such types of questions very carefully. In order to solve similar types of questions, students should follow the same procedure for any composition problem. If ‘ $ f $ ’ and ‘ $ g $ ’ are one-one functions then the composition function $ \left( {fog} \right)\left( x \right) $ is also one-one function. The function composition of two onto function is always onto. Students can find similar types of questions in their NCERT textbooks and practise them for more clarity.
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