WILL GIVE BRAINLIEST PLEASE HELP FAST!! Write the function 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.

Use longhand division or synthetic division. Should help simply.

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oladayoademosu 7 months ago
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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:

https://brainacademy.pro/question/26322830

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