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Sagot :
To determine whether each function is even, odd, or neither, let's start by reviewing the definitions:
- A function [tex]\( f(x) \)[/tex] is even if [tex]\( f(-x) = f(x) \)[/tex].
- A function [tex]\( f(x) \)[/tex] is odd if [tex]\( f(-x) = -f(x) \)[/tex].
- If neither condition is met, the function is neither even nor odd.
Let's examine each function step by step:
1. Function [tex]\( f(x) = \sqrt{x^2} - 9 \)[/tex]:
- Simplify [tex]\( \sqrt{x^2} \)[/tex]. Since [tex]\(\sqrt{x^2} = |x| \)[/tex],
[tex]\[ f(x) = |x| - 9 \][/tex]
- Substitute [tex]\(-x\)[/tex] into [tex]\(f(x)\)[/tex]:
[tex]\[ f(-x) = | -x | - 9 = |x| - 9 \][/tex]
- Notice [tex]\( f(-x) = f(x) \)[/tex], indicating that [tex]\( f(x) \)[/tex] is even.
2. Function [tex]\( g(x) = |x - 3| \)[/tex]:
- Substitute [tex]\(-x\)[/tex] into [tex]\(g(x)\)[/tex]:
[tex]\[ g(-x) = | -x - 3 | = | -(x + 3) | = |x + 3| \][/tex]
- Compare [tex]\( g(-x) \)[/tex] with [tex]\( g(x) \)[/tex]:
[tex]\[ g(x) = | x - 3 | \][/tex]
- Since [tex]\( |x+3| \neq |x-3| \)[/tex], [tex]\( g(x) \)[/tex] is neither even nor odd.
3. Function [tex]\( f(x) = \frac{x}{x^2 - 1} \)[/tex]:
- Substitute [tex]\(-x\)[/tex] into [tex]\(f(x)\)[/tex]:
[tex]\[ f(-x) = \frac{-x}{(-x)^2 - 1} = \frac{-x}{x^2 - 1} \][/tex]
- Compare [tex]\( f(-x) \)[/tex] with [tex]\(-f(x)\)[/tex]:
[tex]\[ -f(x) = -\left( \frac{x}{x^2 - 1} \right) = \frac{-x}{x^2 - 1} \][/tex]
- Since [tex]\( f(-x) = -f(x) \)[/tex], [tex]\( f(x) \)[/tex] is odd.
4. Function [tex]\( g(x) = x + x^2 \)[/tex]:
- Substitute [tex]\(-x\)[/tex] into [tex]\(g(x)\)[/tex]:
[tex]\[ g(-x) = (-x) + (-x)^2 = -x + x^2 \][/tex]
- Compare [tex]\( g(-x) \)[/tex] with [tex]\(g(x)\)[/tex]:
[tex]\[ g(x) = x + x^2 \][/tex]
[tex]\[ g(-x) = -x + x^2 \][/tex]
- Because neither [tex]\( g(-x) = g(x) \)[/tex] nor [tex]\( g(-x) = -g(x) \)[/tex], [tex]\( g(x) \)[/tex] is neither even nor odd.
So we conclude:
- [tex]\( f(x) = \sqrt{x^2} - 9 \)[/tex] is even.
- [tex]\( g(x) = |x - 3| \)[/tex] is neither.
- [tex]\( f(x) = \frac{x}{x^2 - 1} \)[/tex] is odd.
- [tex]\( g(x) = x + x^2 \)[/tex] is neither.
- A function [tex]\( f(x) \)[/tex] is even if [tex]\( f(-x) = f(x) \)[/tex].
- A function [tex]\( f(x) \)[/tex] is odd if [tex]\( f(-x) = -f(x) \)[/tex].
- If neither condition is met, the function is neither even nor odd.
Let's examine each function step by step:
1. Function [tex]\( f(x) = \sqrt{x^2} - 9 \)[/tex]:
- Simplify [tex]\( \sqrt{x^2} \)[/tex]. Since [tex]\(\sqrt{x^2} = |x| \)[/tex],
[tex]\[ f(x) = |x| - 9 \][/tex]
- Substitute [tex]\(-x\)[/tex] into [tex]\(f(x)\)[/tex]:
[tex]\[ f(-x) = | -x | - 9 = |x| - 9 \][/tex]
- Notice [tex]\( f(-x) = f(x) \)[/tex], indicating that [tex]\( f(x) \)[/tex] is even.
2. Function [tex]\( g(x) = |x - 3| \)[/tex]:
- Substitute [tex]\(-x\)[/tex] into [tex]\(g(x)\)[/tex]:
[tex]\[ g(-x) = | -x - 3 | = | -(x + 3) | = |x + 3| \][/tex]
- Compare [tex]\( g(-x) \)[/tex] with [tex]\( g(x) \)[/tex]:
[tex]\[ g(x) = | x - 3 | \][/tex]
- Since [tex]\( |x+3| \neq |x-3| \)[/tex], [tex]\( g(x) \)[/tex] is neither even nor odd.
3. Function [tex]\( f(x) = \frac{x}{x^2 - 1} \)[/tex]:
- Substitute [tex]\(-x\)[/tex] into [tex]\(f(x)\)[/tex]:
[tex]\[ f(-x) = \frac{-x}{(-x)^2 - 1} = \frac{-x}{x^2 - 1} \][/tex]
- Compare [tex]\( f(-x) \)[/tex] with [tex]\(-f(x)\)[/tex]:
[tex]\[ -f(x) = -\left( \frac{x}{x^2 - 1} \right) = \frac{-x}{x^2 - 1} \][/tex]
- Since [tex]\( f(-x) = -f(x) \)[/tex], [tex]\( f(x) \)[/tex] is odd.
4. Function [tex]\( g(x) = x + x^2 \)[/tex]:
- Substitute [tex]\(-x\)[/tex] into [tex]\(g(x)\)[/tex]:
[tex]\[ g(-x) = (-x) + (-x)^2 = -x + x^2 \][/tex]
- Compare [tex]\( g(-x) \)[/tex] with [tex]\(g(x)\)[/tex]:
[tex]\[ g(x) = x + x^2 \][/tex]
[tex]\[ g(-x) = -x + x^2 \][/tex]
- Because neither [tex]\( g(-x) = g(x) \)[/tex] nor [tex]\( g(-x) = -g(x) \)[/tex], [tex]\( g(x) \)[/tex] is neither even nor odd.
So we conclude:
- [tex]\( f(x) = \sqrt{x^2} - 9 \)[/tex] is even.
- [tex]\( g(x) = |x - 3| \)[/tex] is neither.
- [tex]\( f(x) = \frac{x}{x^2 - 1} \)[/tex] is odd.
- [tex]\( g(x) = x + x^2 \)[/tex] is neither.
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