Question

Найти область определения функции​

Answer 1

Ответ:

И снова привет

Объяснение:

1)

$f(x)=\frac{5}{(x-2)(5+x)}$

$f`(x)=\frac{d}{dx} (\frac{5}{(x-2)(5+x)} )$

$f`(x)=\frac{d}{dx} (\frac{5}{5x+x^{2}-10-2x } )$

$f`(x)=\frac{d}{dx} (\frac{5}{3x+x^{2} -10} )$

$f`(x)=-5*\frac{\frac{d}{dx}(3x+x^{2} -10) }{(3x+x^{2} -10)^{2} }$

$f`(x)=-5*\frac{\frac{d}{dx}(3x)+\frac{d}{dx}(x^{2} )-\frac{d}{dx} (10) }{(3x+x^{2} -10)^{2} }$

$f`(x)=-5*\frac{3+\frac{d}{dx} (x^{2} )-\frac{d}{dx}(10) }{(3x+x^{2} -10)^{2} }$

$f`(x)=-5*\frac{3+2x-\frac{d}{dx} (10)}{(3x+x^{2} -10)^{2} }$

$f`(x)=-5*\frac{3+2x-0}{(3x+x^{2} -10)^{2} }$

$f`(x)=-5*\frac{3+2x}{(3x+x^{2} -10)^{2} }$

$f`(x)=-\frac{15+10x}{(x^{2} +3x-10)^{2} }$

2)

$f(x)=\frac{7}{x^{2} +6x+9} \\\\f`(x)=\frac{d}{dx} (\frac{7}{x^{2} +6x+9} )\\\\f`(x)=\frac{d}{dx} (\frac{7}{x^{2} +6x+9} )\\\\f`(x)=-7*\frac{\frac{d}{dx}(x^{2} +6x+9) }{(x^{2} +6x+9)^{2} } \\\\f`(x)=-7*\frac{\frac{d}{dx}(x^{2} )+\frac{d}{dx}(6x)+\frac{d}{dx} (9) }{(x^{2} +6x+9)^{2} } \\f`(x)=-7*\frac{2x+6}{(x^{2} +6x+9)^{2} }$

$f`(x)=-\frac{14}{(x+3)^{3} }$

3)$f(x)=\sqrt{x+5} +\sqrt{x-2} \\y=\sqrt{x+5} +\sqrt{x-2} \\x=\sqrt{y+5} +\sqrt{y-2} \\\sqrt{y+5} +\sqrt{y-2} =x\\\sqrt{y+5}=x-\sqrt{y-2} \\y+5=x^{2} -2x\sqrt{y-2} +y-2\\5=x^{2} -2x\sqrt{y-2} -2\\2x\sqrt{y-2} =x^{2} -2-5\\2x\sqrt{y-2} =x^{2} -7\\\sqrt{y-2} =\frac{x^{2}-7 }{2x} \\y-2=\frac{(x^{2} -7)^{2} }{4x^{2} } \\y-2=\frac{x^{4}-14x^{2} +49 }{4x^{2} } \\y=\frac{x^{4}-14x^{2} +49 }{4x^{2} } \\f^{-1} (x)=\frac{x^{4}-14x^{2} +49 }{4x^{2} } +2$

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