Consider the function \(f(x)\), \(g(x)\), \(h(x)\) as given below. Verify that \((f \circ g) \circ h = f \circ (g \circ h)\) in each case.
(i) \(f(x) = x^2\), \(g(x) = 2x\) and \(h(x) = x + 5\)
(ii) \(f(x) = x - 5\), \(g(x) = x^2\) and \(h(x) = 3x - 5\)
Proof:
(i) \(f \circ g(x) =\)
\((f \circ g) \circ h =\)
\(g \circ h(x) =\)
\(f \circ (g \circ h) =\)
Thus,
(ii) \(f \circ g(x) =\)
\((f \circ g) \circ h =\)
\(g \circ h(x) =\)
\(f \circ (g \circ h) =\)
Thus,
Answer variants:
\((f \circ g) \circ h \neq f \circ (g \circ h)\)
\(4x^2\)
\((f \circ g) \circ h = f \circ (g \circ h)\)
\(9x^2 - 30x + 25\)