Докажите, что функция y=f(x) четная. [tex]a) f(x) =7cos4x+3 x^{2} \\\\b) f (x) = \frac{x^{2}-x}{x+2} - \frac{x^{2}+x}{x-2}[/tex]. Created by DarkPh0enix. algebra-ru

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July 2022 2 15 Report

Докажите, что функция y=f(x) четная.

[tex]a) f(x) =7cos4x+3 x^{2} \\\\b) f (x) = \frac{x^{2}-x}{x+2} - \frac{x^{2}+x}{x-2}[/tex]

AntVa

a) f(-x) =7cos(-4x)+3(-x)²=7cos4x+3x²=f(x); косинус четная функция!

b) $f(-x)=\frac{(-x)^2+x}{-x+2}-\frac{(-x)^2-x}{-x-2}=\frac{x^2+x}{-(x-2)}-\frac{x^2-x}{-(x+2)}=-\frac{x^2+x}{x-2}+\frac{x^2-x}{x+2}=\frac{x^2-x}{x+2}-\frac{x^2+x}{x-2}=f(x)$

2 votes Thanks 1

spasibo3pajbrh
$f(x) =7cos4x+3 x^{2} \\ f( - x) = \\ = 7cos( 4\times (- x))+3 ( - x)^{2} = \\ = 7cos( - 4x) + 3 {x}^{2} = \\ =7cos4x+3 x^{2} = f(x)$
$f(x) = \frac{x^{2}-x}{x+2} - \frac{x^{2}+x}{x-2} = \\ = \frac{( - x) ^{2} - ( - x) }{ - x + 2} - \frac{( - x) ^{2} + ( - x)}{ - x - 2} = \\ = - \frac{ {x}^{2} + x }{(x - 2)} + \frac{ {x }^{2} - x }{x + 2} = \\ = \frac{x^{2}-x}{x+2} - \frac{x^{2}+x}{x-2} = f(x)$

1 votes Thanks 1

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