Question

Investigate the family of functions \({f_n}\left( x \right) = tanh\left( {nsinx} \right)\)where \(n\) is a positive integer. Describe what happens to the graph of \({f_n}\) when \(n\) becomes large.

Short answer

As \(n\) becomes large the graph of \({f_n}\left( x \right)\) approaches the graph of \(y = 1\) on the intervals \(\left( {2k\pi ,\left( {2k + 1} \right)\pi } \right)\) and \(y = - 1\) on the intervals \(\left( {\left( {2k - 1} \right)\pi ,2k\pi } \right)\).

Step-by-step solution

Step 1: Construct the graph of function

Describe what happens to the graph of \({f_n}\) when \(n\) becomes large

Consider the function \({f_n}\left( x \right) = \tanh \left( {n\sin x} \right)\) where \(n\) is a positive integer.

From figure \(3, - 1 < \tanh x < 1\) so we can choose a viewing rectangle of \(\left( {0,2\pi } \right) \times \left( { - 1,1} \right)\).

From the graph it can be seen that \({f_n}\left( x \right)\) becomes more rectangular looking as \(n\) becomes large.

Thus, as \(n\) becomes large the graph of \({f_n}\left( x \right)\) approaches the graph of \(y = 1\) on the intervals \(\left( {2k\pi ,\left( {2k + 1} \right)\pi } \right)\) and \(y = - 1\) on the intervals \(\left( {\left( {2k - 1} \right)\pi ,2k\pi } \right)\).