Решите срочно математику, пожалуйста!!!. Created by okc0677. algebra-ru

Решите срочно математику, пожалуйста!!!...

okc0677 @okc0677

July 2022 1 12 Report

Решите срочно математику, пожалуйста!!!

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Verified answer

$f(x)=\dfrac{x+3}{x^{2}+1 } \\\\f'(x)=\Big(\dfrac{x+3}{x^{2}+1 }\Big)'=\dfrac{(x+3)'\cdot(x^{2}+1)-(x+3)\cdot(x^{2} +1)' }{(x^{2}+1)^{2}} =\\\\=\dfrac{x^{2}+1-(x+3)\cdot 2x }{(x^{2}+1)^{2}}=\dfrac{x^{2}+1-2x^{2}-6x }{(x^{2}+1)^{2} } =\dfrac{-x^{2}-6x+1 }{(x^{2}+1)^{2}}\\\\\\\dfrac{-x^{2}-6x+1 }{(x^{2}+1)^{2}}=0\\\\\left\{\begin{array}{ccc}-x^{2} -6x+1=0\\x^{2}+1\neq0 \end{array}\right\\\\\\x^{2}+6x-1=0\\\\D=6^{2}-4\cdot (-1)=36+4=40=(2\sqrt{10})^{2}$

$x_{1} =\dfrac{-6-2\sqrt{10} }{2} =-3-\sqrt{10} \\\\x_{2} =\dfrac{-6+2\sqrt{10} }{2} =\sqrt{10}-3$

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[-7]_____ - 3 - √10_______ √10 - 3____[7]

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x ∈ [- 7 ; - 3 - √10] ∪ [√10 - 3 ; 7]

Ответ : 8

1 votes Thanks 2

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Verified answer

113e75a115418980ab8625a74471f8d6

66451005ab3b1a17c832afd3689abd06

2 637b1cdc767bf

16610d5940c2e409cc5f281838a94a0b

3933dc04e96bf71f2d812218fa2814b0

05610467c2fc5c3f53c24eb2b9420e78

e441c07646fbddf453e148adff2d4314

f1633dde53b0b8d27c2d72643fcb25ab

30 6376da8886181

928bd52eefb2e833cd179effb876526b

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