f(x) = (2x³ + 1 + x²)/(x² + 9) as a sum of two functions g(x) and h(x), where g is an odd function and h is an even function is f(x) = 2x³/(x² + 9) + (1 + x²)/(x² + 9).
To answer the questions, we need to know what functions are
What are functions?
A function is a mathematical expression that shows the relationship between two variables.
Types of function
- Odd functions - These are functions in which f(-x) = -f(x)
- Even functions - These are functions in which f(-x) = f(x)
- Neither - When the function is neither even nor odd.
Since f(x) = (2x³ + 1 + x²)/(x² + 9).
The odd part of f(x)
Its odd part g(x) = [f(x) - f(-x)]/2
So, g(x) = [f(x) - f(-x)]/2
= [(2x³ + 1 + x²)/(x² + 9) - (2(-x)³ + 1 + (-x)²)/((-x)² + 9)]/2
= [(2x³ + 1 + x²)/(x² + 9) - (-2x³ + 1 + x²)/(x² + 9)]/2
= (2x³ + 1 + x² + 2x³ - 1 - x²)/2(x² + 9)
= 4x³/2(x² + 9)
g(x) = 2x³/(x² + 9)
The even part of f(x)
Its even part h(x) = [f(x) + f(-x)]/2
So, h(x) = [f(x) + f(-x)]/2
= [(2x³ + 1 + x²)/(x² + 9) + (2(-x)³ + 1 + (-x)²)/((-x)² + 9)]/2
= [(2x³ + 1 + x²)/(x² + 9) + (-2x³ + 1 + x²)/(x² + 9)]/2
= (2x³ + 1 + x² - 2x³ + 1 + x²)/2(x² + 9)
= 2(1 + x²)/2(x² + 9)
h(x) = (1 + x²)/(x² + 9)
So, f(x) = g(x) + h(x)
f(x) = 2x³/(x² + 9) + (1 + x²)/(x² + 9)
So, f(x) = (2x³ + 1 + x²)/(x² + 9) as a sum of two functions g(x) and h(x), where g is an odd function and h is an even function is f(x) = 2x³/(x² + 9) + (1 + x²)/(x² + 9).
Learn more about odd and even functions here:
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