
If \[f\left( x \right)\] is symmetric about the line \[x=a\] and \[x=b\], where \[a< b\] then find the period of \[f\left( x \right)\].
Answer
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Hint: Since we are given with \[f\left( x \right)\] is symmetric about the lines \[x=a\] and \[x=b\], then we must substitute these values into the function and compute them. Upon computing them we obtain the period of the function. This would be our required answer.
Complete step by step solution:
Now let us learn about the period of a function. The distance between the repetition of any function is called the period of the function. In the case of the trigonometric function, the length of one complete cycle is called a period. Each function will have its own period. To any function, the reciprocal of the period is called the frequency of the function. We can find a period when it is represented as \[f\left( x \right)=f\left( x+p \right)\], \[p\] is the period of the function.
Now let us find the period of \[f\left( x \right)\] which is symmetric about the line \[x=a\] and \[x=b\].
Since \[f\left( x \right)\] is symmetric about the lines,
We have \[x=a\] and \[x=b\].
Then,
\[\begin{align}
& f\left( a-x \right)=f\left( a+x \right) \\
& f\left( b-x \right)=f\left( b+x \right) \\
\end{align}\]
Now we have,
\[\begin{align}
& f\left( x \right)=f\left( a+\left( x-a \right) \right) = f\left( a-\left( x-a \right) \right) = f\left( 2a-x \right) \\
& \Rightarrow f\left( b+\left( 2a-x-b \right) \right) = f\left( b-\left( 2a-x-b \right) \right) \\
& = f\left( 2b-2a+x \right) \\
&= f\left( x+\left( 2b-2a \right) \right) \\
\end{align}\]
\[\therefore \] Generally, the period of a function can be determined as \[f\left( x \right)=f\left( x+p \right)\] in which \[p\] is the period of the function.
So we can conclude that the period of \[f\left( x \right)\] is \[\left( 2b-2a \right)\].
Note: While computing for the period, we must consider the given conditions and substitute into the function. We must also note that the period repeats at a particular interval of time. This would be termed as periodicity. We can represent the period of a trigonometric function in a graph. We must also note that the cotangent and the tangent functions are periodic but they do have breaks in the graph.
Complete step by step solution:
Now let us learn about the period of a function. The distance between the repetition of any function is called the period of the function. In the case of the trigonometric function, the length of one complete cycle is called a period. Each function will have its own period. To any function, the reciprocal of the period is called the frequency of the function. We can find a period when it is represented as \[f\left( x \right)=f\left( x+p \right)\], \[p\] is the period of the function.
Now let us find the period of \[f\left( x \right)\] which is symmetric about the line \[x=a\] and \[x=b\].
Since \[f\left( x \right)\] is symmetric about the lines,
We have \[x=a\] and \[x=b\].
Then,
\[\begin{align}
& f\left( a-x \right)=f\left( a+x \right) \\
& f\left( b-x \right)=f\left( b+x \right) \\
\end{align}\]
Now we have,
\[\begin{align}
& f\left( x \right)=f\left( a+\left( x-a \right) \right) = f\left( a-\left( x-a \right) \right) = f\left( 2a-x \right) \\
& \Rightarrow f\left( b+\left( 2a-x-b \right) \right) = f\left( b-\left( 2a-x-b \right) \right) \\
& = f\left( 2b-2a+x \right) \\
&= f\left( x+\left( 2b-2a \right) \right) \\
\end{align}\]
\[\therefore \] Generally, the period of a function can be determined as \[f\left( x \right)=f\left( x+p \right)\] in which \[p\] is the period of the function.
So we can conclude that the period of \[f\left( x \right)\] is \[\left( 2b-2a \right)\].
Note: While computing for the period, we must consider the given conditions and substitute into the function. We must also note that the period repeats at a particular interval of time. This would be termed as periodicity. We can represent the period of a trigonometric function in a graph. We must also note that the cotangent and the tangent functions are periodic but they do have breaks in the graph.
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