Math Problem Statement

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Solution

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14. Diketahui f1(x)=xx+2f^{-1}(x) = \frac{x}{x+2} dan g1(x)=23xg^{-1}(x) = \frac{2}{3x}. Tentukan nilai (fg)(2)(f \circ g)(2).

To solve this, we need to find the composition of the functions f1(x)f^{-1}(x) and g1(x)g^{-1}(x) and evaluate it at x=2x = 2.

  1. Step 1: Find g1(2)g^{-1}(2). g1(x)=23xg^{-1}(x) = \frac{2}{3x} Substituting x=2x = 2: g1(2)=23(2)=26=13g^{-1}(2) = \frac{2}{3(2)} = \frac{2}{6} = \frac{1}{3}

  2. Step 2: Now find f1(g1(2))=f1(13)f^{-1}(g^{-1}(2)) = f^{-1}\left(\frac{1}{3}\right). f1(x)=xx+2f^{-1}(x) = \frac{x}{x+2} Substituting x=13x = \frac{1}{3}: f1(13)=1313+2=1373=17f^{-1}\left(\frac{1}{3}\right) = \frac{\frac{1}{3}}{\frac{1}{3} + 2} = \frac{\frac{1}{3}}{\frac{7}{3}} = \frac{1}{7}

Thus, (fg)(2)=17(f \circ g)(2) = \frac{1}{7}.

15. Diketahui f(x)=2x5f(x) = 2x - 5, g1(x)=x+34g^{-1}(x) = \frac{x+3}{4}, dan h1(x)=8xh^{-1}(x) = 8x. Tentukan nilai (fgh)1(7)(f \circ g \circ h)^{-1}(-7).

This question asks us to solve for the inverse of the composition (fgh)1(7)(f \circ g \circ h)^{-1}(-7).

  1. Step 1: Find h1(7)h^{-1}(-7). h1(x)=8xh^{-1}(x) = 8x Substituting x=7x = -7: h1(7)=8(7)=56h^{-1}(-7) = 8(-7) = -56

  2. Step 2: Find g1(h1(7))=g1(56)g^{-1}(h^{-1}(-7)) = g^{-1}(-56). g1(x)=x+34g^{-1}(x) = \frac{x + 3}{4} Substituting x=56x = -56: g1(56)=56+34=534=534g^{-1}(-56) = \frac{-56 + 3}{4} = \frac{-53}{4} = -\frac{53}{4}

  3. Step 3: Find f(g1(h1(7)))=f(534)f(g^{-1}(h^{-1}(-7))) = f\left(-\frac{53}{4}\right). f(x)=2x5f(x) = 2x - 5 Substituting x=534x = -\frac{53}{4}: f(534)=2(534)5=10645=1064204=1264=632f\left(-\frac{53}{4}\right) = 2\left(-\frac{53}{4}\right) - 5 = -\frac{106}{4} - 5 = -\frac{106}{4} - \frac{20}{4} = -\frac{126}{4} = -\frac{63}{2}

Thus, (fgh)1(7)=632(f \circ g \circ h)^{-1}(-7) = -\frac{63}{2}.

16. Diketahui f(x3)=6x11f(x-3) = 6x - 11 dan g(x+1)=2x63x+4g(x+1) = \frac{2x - 6}{3x + 4}. Tentukan komposisi fungsi (fg)(2a+3)(f \circ g)(2a + 3).

To find (fg)(2a+3)(f \circ g)(2a + 3), we need to substitute the expression 2a+32a + 3 into g(x)g(x), then apply f(x3)f(x-3).

  1. Step 1: Find g(2a+3+1)=g(2a+4)g(2a + 3 + 1) = g(2a + 4). g(x+1)=2x63x+4g(x+1) = \frac{2x - 6}{3x + 4} Substituting x=2a+3x = 2a + 3: g(2a+4)=2(2a+4)63(2a+4)+4=4a+866a+12+4=4a+26a+16g(2a + 4) = \frac{2(2a+4) - 6}{3(2a+4) + 4} = \frac{4a + 8 - 6}{6a + 12 + 4} = \frac{4a + 2}{6a + 16}

  2. Step 2: Now substitute into f(x3)f(x - 3), where x=4a+26a+16x = \frac{4a + 2}{6a + 16}.

Thus, the result is f(4a+26a+163)f \left( \frac{4a + 2}{6a + 16} - 3 \right). To simplify further, we would expand and evaluate.


Would you like further clarification on these solutions? Here are five related questions to deepen your understanding:

  1. How do you find the composition of two functions?
  2. What are the key steps to solve for the inverse of a function?
  3. Why is simplifying complex fractions important in function composition?
  4. What role do inverses play in solving composite function problems?
  5. How can transformations of functions (like shifting by constants) affect their evaluation?

Tip: When dealing with composite functions, always evaluate the inner functions first and work your way outwards!

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Math Problem Analysis

Mathematical Concepts

Inverse Functions
Function Composition
Algebraic Manipulation

Formulas

f^{-1}(x) = \frac{x}{x+2}
g^{-1}(x) = \frac{2}{3x}
f(x) = 2x - 5
g^{-1}(x) = \frac{x+3}{4}
h^{-1}(x) = 8x

Theorems

Inverse Function Theorem
Function Composition

Suitable Grade Level

Grades 10-12