Function Composition Practice

Composition of Functions from Tables and Graphs

  • Table Evaluations:

    • Given discrete function values from a table:
    • (f∘g)(1)=f(g(1))=f(0)=5(f \circ g)(1) = f(g(1)) = f(0) = 5
    • (g∘f)(2)=g(f(2))=g(1)=0(g \circ f)(2) = g(f(2)) = g(1) = 0
    • (g∘g)(1)=g(g(1))=g(0)=1(g \circ g)(1) = g(g(1)) = g(0) = 1
  • Graphical Evaluations:   Graph of f(x) and g(x)

    • (g∘f)(5)=g(f(5))=g(1)=4(g \circ f)(5) = g(f(5)) = g(1) = 4
    • (f∘g)(2)=f(g(2))=f(2)=−2(f \circ g)(2) = f(g(2)) = f(2) = -2

Function Decomposition

  • Decomposing Composite Functions H(x)=(f∘g)(x)H(x) = (f \circ g)(x):
    • For H(x)=(2x+3)4H(x) = (2x + 3)^4: g(x)=2x+3g(x) = 2x + 3 and f(x)=x4f(x) = x^4
    • For H(x)=(1+x2)3H(x) = (1 + x^2)^3: g(x)=x2g(x) = x^2 and f(x)=(1+x)3f(x) = (1 + x)^3
    • For H(x)=∣2x2+3∣H(x) = |2x^2 + 3|: g(x)=2x2+3g(x) = 2x^2 + 3 and f(x)=∣x∣f(x) = |x|

Algebraic Combinations and Compositions

  • Given Base Functions:

    • f(x)=x2f(x) = x^2
    • g(x)=1xg(x) = \frac{1}{x}
    • h(x)=xh(x) = \sqrt{x}
    • k(x)=exk(x) = e^x
    • A(x)=ln⁡(x)A(x) = \ln(x)
  • Expressions via Combinations and Compositions:

    • ex=(k∘h)(x)e^{\sqrt{x}} = (k \circ h)(x)
    • x2ex=f(x)⋅k(x)x^2 e^x = f(x) \cdot k(x)
    • exx2=k(x)f(x)\frac{e^x}{x^2} = \frac{k(x)}{f(x)}
    • x2+x=f(x)+h(x)x^2 + \sqrt{x} = f(x) + h(x)
    • ln⁡(1x)=(A∘g)(x)\ln\left(\frac{1}{x}\right) = (A \circ g)(x)
    • ln⁡(x)−ex=A(x)−k(x)\ln(x) - e^x = A(x) - k(x)

Compositions and Domain Restrictions

  • Polynomial Compositions:

    • Given f(x)=x2f(x) = x^2 and g(x)=x2−4g(x) = x^2 - 4:
    • (f∘g)(x)=(x2−4)2=x4−8x2+16(f \circ g)(x) = (x^2 - 4)^2 = x^4 - 8x^2 + 16 with domain (−∞,∞)(-\infty, \infty)
    • (g∘f)(x)=(x2)2−4=x4−4(g \circ f)(x) = (x^2)^2 - 4 = x^4 - 4 with domain (−∞,∞)(-\infty, \infty)
  • Rational Compositions:

    • Given f(x)=1x+3f(x) = \frac{1}{x + 3} and g(x)=−2xg(x) = -\frac{2}{x}:
    • (f∘g)(x)=x3x−2(f \circ g)(x) = \frac{x}{3x - 2} with domain (−∞,0)∪(0,23)∪(23,∞)(-\infty, 0) \cup \left(0, \frac{2}{3}\right) \cup \left(\frac{2}{3}, \infty\right)
    • (g∘f)(x)=−2(x+3)=−2x−6(g \circ f)(x) = -2(x + 3) = -2x - 6 with domain (−∞,−3)∪(−3,∞)(-\infty, -3) \cup (-3, \infty)