Math Problem Statement

Mid Semester Exam (Mathematics, Grade XI, 2024/2025): Solve the following based on the functions f(x) = 3x + 2, g(x) = 5x - 4, h(x) = x^2. Questions include addition, subtraction, multiplication of functions, composition of functions, and finding inverses.

Solution

Let's break down the math problems based on the given image.

We are provided with the following functions:

  • f(x)=3x+2f(x) = 3x + 2
  • g(x)=5x4g(x) = 5x - 4
  • h(x)=x2h(x) = x^2

The problems to solve:

  1. f(x)+g(x)f(x) + g(x)
  2. f(x)g(x)f(x) - g(x)
  3. f(x)g(x)f(x) \cdot g(x)
  4. (fg)(x)(f \circ g)(x) (This means f(g(x))f(g(x)))
  5. (gf)(x)(g \circ f)(x) (This means g(f(x))g(f(x)))
  6. (gh)(x)(g \circ h)(x)
  7. (fg1)(x)(f \circ g^{-1})(x) (This involves the inverse of g(x)g(x))
  8. (gf)(2)(g \circ f)(2) (Evaluate g(f(2))g(f(2)))
  9. f1(x)f^{-1}(x) (The inverse of f(x)f(x))
  10. (g1h)(x)(g^{-1} \circ h)(x) (This involves g1(x)g^{-1}(x) and applying h(x)h(x))

I'll now solve these step-by-step:

Solutions:

  1. f(x)+g(x)=(3x+2)+(5x4)=8x2f(x) + g(x) = (3x + 2) + (5x - 4) = 8x - 2

  2. f(x)g(x)=(3x+2)(5x4)=3x+25x+4=2x+6f(x) - g(x) = (3x + 2) - (5x - 4) = 3x + 2 - 5x + 4 = -2x + 6

  3. f(x)g(x)=(3x+2)(5x4)=15x212x+10x8=15x22x8f(x) \cdot g(x) = (3x + 2)(5x - 4) = 15x^2 - 12x + 10x - 8 = 15x^2 - 2x - 8

  4. (fg)(x)=f(g(x))=f(5x4)=3(5x4)+2=15x12+2=15x10(f \circ g)(x) = f(g(x)) = f(5x - 4) = 3(5x - 4) + 2 = 15x - 12 + 2 = 15x - 10

  5. (gf)(x)=g(f(x))=g(3x+2)=5(3x+2)4=15x+104=15x+6(g \circ f)(x) = g(f(x)) = g(3x + 2) = 5(3x + 2) - 4 = 15x + 10 - 4 = 15x + 6

  6. (gh)(x)=g(h(x))=g(x2)=5x24(g \circ h)(x) = g(h(x)) = g(x^2) = 5x^2 - 4

  7. (fg1)(x)(f \circ g^{-1})(x): To do this, we first need g1(x)g^{-1}(x).

    • y=5x4y = 5x - 4 → Solve for xx: y+4=5xy + 4 = 5x, x=y+45x = \frac{y + 4}{5}
    • So, g1(x)=x+45g^{-1}(x) = \frac{x + 4}{5}.
    • Now (fg1)(x)=f(x+45)=3(x+45)+2=3(x+4)5+2=3x+125+2=3x+125+105=3x+225(f \circ g^{-1})(x) = f\left(\frac{x + 4}{5}\right) = 3\left(\frac{x + 4}{5}\right) + 2 = \frac{3(x + 4)}{5} + 2 = \frac{3x + 12}{5} + 2 = \frac{3x + 12}{5} + \frac{10}{5} = \frac{3x + 22}{5}
  8. (gf)(2)(g \circ f)(2): First, find f(2)f(2), then apply gg.

    • f(2)=3(2)+2=6+2=8f(2) = 3(2) + 2 = 6 + 2 = 8
    • g(8)=5(8)4=404=36g(8) = 5(8) - 4 = 40 - 4 = 36
  9. f1(x)f^{-1}(x): To find the inverse of f(x)=3x+2f(x) = 3x + 2, solve for xx.

    • y=3x+2y = 3x + 2
    • y2=3xy - 2 = 3x
    • x=y23x = \frac{y - 2}{3}
    • So, f1(x)=x23f^{-1}(x) = \frac{x - 2}{3}
  10. (g1h)(x)(g^{-1} \circ h)(x): First apply h(x)h(x), then g1g^{-1}.

  • h(x)=x2h(x) = x^2
  • Now, apply g1(x2)=x2+45g^{-1}(x^2) = \frac{x^2 + 4}{5}

Let me know if you need more details or have any questions.

Here are 5 additional questions to extend your understanding:

  1. What is the difference between the composition of functions and the sum of functions?
  2. How do you derive the inverse of a linear function like f(x)=3x+2f(x) = 3x + 2?
  3. How does composition affect the domain and range of functions like f(x)f(x) and g(x)g(x)?
  4. What are the implications of applying the inverse of a function before another function in composition?
  5. How can you check if two functions are inverses of each other?

Tip: Always remember to apply functions in the correct order when performing compositions, especially when inverses are involved.

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Math Problem Analysis

Mathematical Concepts

Function Operations
Function Composition
Inverse Functions
Algebra

Formulas

f(x) = 3x + 2
g(x) = 5x - 4
h(x) = x^2
(f + g)(x) = f(x) + g(x)
(f - g)(x) = f(x) - g(x)
(f * g)(x) = f(x) * g(x)
(f ∘ g)(x) = f(g(x))
f⁻¹(x) = (x - 2)/3
g⁻¹(x) = (x + 4)/5

Theorems

Inverse of a Function
Composition of Functions
Function Operations

Suitable Grade Level

Grade 11