Notes on Decomposing h(x) into f∘g

Overview of Composite Function Decomposition

  • Goal: Find functions f and g such that the composition f ∘ g equals a given function h.
  • In the example, the target is: h(x)=(3x2−5)21.h(x) = \big(3x^{2} - 5\big)^{21}.
  • Key idea: If you can express (h(x)) as (F\big(G(x)\big)), then set (g(x) = G(x)) and (f(x) = F(x)) so that (f \circ g = h).

Given Function h

  • The transcript states: the function h raises the inner expression (3x^{2} - 5) to the 21st power, i.e.,
  • h(x)=(3x2−5)21.h(x) = \big(3x^{2} - 5\big)^{21}.
  • This reveals an inner expression and an outer operation, suggesting a natural decomposition.

Decomposition Strategy

  • Recognize inner expression (the argument of the outer operation) as the input to the outer function.
  • If h(x) has the form (h(x) = (u(x))^{21}) then a natural choice is:
  • Outer function: (f(t) = t^{21}).
  • Inner function: (g(x) = u(x)).
  • Then the composition (f\circ g) yields the target: (f(g(x)) = \big(u(x)\big)^{21} = h(x)).

Worked Example

  • Identify inner expression: (u(x) = 3x^{2} - 5).
  • Define the outer function: (f(t) = t^{21}).
  • Define the inner function: (g(x) = 3x^{2} - 5).
  • Express f and g:
  • f(x)=x21.f(x) = x^{21}.
  • g(x)=3x2−5.g(x) = 3x^{2} - 5.
  • Compute the composition:
  • f∘g(x)=f(g(x))=f(3x2−5).f\circ g (x) = f(g(x)) = f(3x^{2} - 5).
  • Apply the rule of f:
  • f(g(x))=(3x2−5)21.f(g(x)) = (3x^{2} - 5)^{21}.
  • This is exactly (h(x)), i.e., h(x)=(3x2−5)21.h(x) = (3x^{2} - 5)^{21}.

Verification

  • By construction, (f\circ g) reproduces (h):
  • f∘g(x)=(3x2−5)21=h(x).f\circ g (x) = (3x^{2} - 5)^{21} = h(x).
  • Therefore, the chosen (f) and (g) satisfy (f\circ g = h).

Takeaways and Implications

  • When given a function of the form (h(x) = (u(x))^{n}), a standard decomposition is to set (f(t) = t^{n}) and (g(x) = u(x)).
  • This provides a straightforward method to identify components of a composite function from its algebraic form.
  • The approach mirrors how the outer and inner functions interact under composition: the inner expression (u(x)) becomes the input to the outer function (f).
  • This technique is foundational for understanding more complex compositions and sets up intuition for methods like the chain rule in calculus, where recognizing inner and outer layers matters for differentiation.