
Given that \[f\] is even function and \[g\] is odd function, how do you determine whether \[h\] is even or odd or neither \[h\left( x \right) = 2f\left( x \right) + xg\left( x \right)\] ?
Answer
533.7k+ views
Hint: In order to solve the above question, you need to have a basic understanding of function composition. In an informal language we can say that “it is a function of a function”. This refers to when we use one function as an argument of the other function. You need to remember the concepts of odd and even functions to solve this question.
Complete step by step solution:
We are given that \[f\] is an even function, so it can be written as: \[f\left( { - x} \right) = f\left( x \right)\] .
We are also given that \[g\] is an odd function, this can be written as: \[g\left( { - x} \right) = - g\left( x \right)\] .
Now, we have: \[h\left( x \right) = 2f\left( x \right) + xg\left( x \right)\] .
We can write this as:
Changing \[x = - x\] , we get,
\[h\left( { - x} \right) = 2f\left( { - x} \right) + \left( { - x} \right)g\left( { - x} \right)\] .
Now we know that since \[f\] is an even function, so \[f\left( { - x} \right) = f\left( x \right)\] and \[g\] is odd function, so \[g\left( { - x} \right) = - g\left( x \right)\] . Using these values in the above equation we get,
\[h\left( { - x} \right) = 2f\left( x \right) + \left( { - x} \right)\left( { - g\left( x \right)} \right)\] .
\[ = 2f\left( x \right) + xg\left( x \right)\] .
Now, we are given the question that \[h\left( x \right) = 2f\left( x \right) + xg\left( x \right)\] . So, putting this value of \[h\left( x \right)\] in the above statement, we get,
\[h\left( { - x} \right) = h\left( x \right)\] .
So, from this we understand that \[h\] is an even function.
Hence, our final answer is that \[h\] is an even function.
So, the correct answer is “ \[h\] is an even function”.
Note: In order to solve questions similar to the one given above, always remember the concept of odd and even functions. In this question, we used that even function can be written as \[f\left( { - x} \right) = f\left( x \right)\] whereas odd functions can be written as \[g\left( { - x} \right) = - g\left( x \right)\] . These simple principles will help you in solving your questions.
Complete step by step solution:
We are given that \[f\] is an even function, so it can be written as: \[f\left( { - x} \right) = f\left( x \right)\] .
We are also given that \[g\] is an odd function, this can be written as: \[g\left( { - x} \right) = - g\left( x \right)\] .
Now, we have: \[h\left( x \right) = 2f\left( x \right) + xg\left( x \right)\] .
We can write this as:
Changing \[x = - x\] , we get,
\[h\left( { - x} \right) = 2f\left( { - x} \right) + \left( { - x} \right)g\left( { - x} \right)\] .
Now we know that since \[f\] is an even function, so \[f\left( { - x} \right) = f\left( x \right)\] and \[g\] is odd function, so \[g\left( { - x} \right) = - g\left( x \right)\] . Using these values in the above equation we get,
\[h\left( { - x} \right) = 2f\left( x \right) + \left( { - x} \right)\left( { - g\left( x \right)} \right)\] .
\[ = 2f\left( x \right) + xg\left( x \right)\] .
Now, we are given the question that \[h\left( x \right) = 2f\left( x \right) + xg\left( x \right)\] . So, putting this value of \[h\left( x \right)\] in the above statement, we get,
\[h\left( { - x} \right) = h\left( x \right)\] .
So, from this we understand that \[h\] is an even function.
Hence, our final answer is that \[h\] is an even function.
So, the correct answer is “ \[h\] is an even function”.
Note: In order to solve questions similar to the one given above, always remember the concept of odd and even functions. In this question, we used that even function can be written as \[f\left( { - x} \right) = f\left( x \right)\] whereas odd functions can be written as \[g\left( { - x} \right) = - g\left( x \right)\] . These simple principles will help you in solving your questions.
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