Consider the function \(f(u)\), \(g(u)\), \(h(u)\) as given below. Show that \((f \circ g) \circ h = f \circ (g \circ h)\) in the below case.
\(f(u) = u - 1\), \(g(u) = 3u + 1\) and \(h(u) = u^2\)
Proof:
\(f \circ g\) \(=\)
\((f \circ g) \circ h\) \(=\) - - - - - - - (I)
\(g \circ h\) \(=\)
\(f \circ (g \circ h)\) \(=\) - - - - - - - (II)
Equation (I) Equation (II)
Hence, proved.