Ответ:
Объяснение:
[tex]f(x)=4x-x^2-3.[/tex]
[tex]4x-x^2-3=0\ |*(-1)\\\\x^2-4x+3=0\\\\x^2-x-3x+3=0\\\\x*(x-1)-3*(x-1)=0\\\\(x-1)*(x-3)=0\\\\x-1=0\\\\x_1=1.\\\\x-3=0\\\\x_2=3.[/tex]
1) x₀=1.
[tex]f(1)=4*1-1^2-3=4-1-3=0.\\\\f'(x)=(4x-x^2-3)'=4-2x.\\\\f'(1)=4-2*1=2.\ \ \ \ \Rightarrow\\\\f(1)_k=0+2*(x-1)=2x-2.[/tex]
2) x₀=3.
[tex]f(3)=4*3-3^3-3=12-9-3=0.\\\\f'(x)=4-2x\\\\f'(3)=4-2*3=-2\ \ \ \ \Rightarrow\\\\f(3)_k=0+(-2)*(x-3)=-2x+6.[/tex]
Copyright © 2024 SCHOLAR.TIPS - All rights reserved.
Answers & Comments
Ответ:
Объяснение:
[tex]f(x)=4x-x^2-3.[/tex]
[tex]4x-x^2-3=0\ |*(-1)\\\\x^2-4x+3=0\\\\x^2-x-3x+3=0\\\\x*(x-1)-3*(x-1)=0\\\\(x-1)*(x-3)=0\\\\x-1=0\\\\x_1=1.\\\\x-3=0\\\\x_2=3.[/tex]
1) x₀=1.
[tex]f(1)=4*1-1^2-3=4-1-3=0.\\\\f'(x)=(4x-x^2-3)'=4-2x.\\\\f'(1)=4-2*1=2.\ \ \ \ \Rightarrow\\\\f(1)_k=0+2*(x-1)=2x-2.[/tex]
2) x₀=3.
[tex]f(3)=4*3-3^3-3=12-9-3=0.\\\\f'(x)=4-2x\\\\f'(3)=4-2*3=-2\ \ \ \ \Rightarrow\\\\f(3)_k=0+(-2)*(x-3)=-2x+6.[/tex]