[m]f(x)=\sqrt{\frac{4-x}{1+x}}[/m]
x_(o)=3
x_(o)+ Δx=3,2
Δx=0,2
f(x_(o)+ Δx)-f(x_(o)) ≈ f`(x_(o))* Δx ⇒ [b]f(x_(o)+ Δx) ≈ f(x_(o)) + f`(x_(o))* Δx [/b]
[m]f(3)=\sqrt{\frac{4-3}{1+3}}=\frac{1}{2}[/m]
[m]f`(x)=(\sqrt{\frac{4-x}{1+x}})`=\frac{1}{2\sqrt{\frac{4-x}{1+x}}}\cdot (\frac{4-x}{1+x})`=-\frac{5}{2\sqrt{(4-x)\cdot (1+x)^3}}[/m]
[m]f`(3)=-\frac{5}{2\sqrt{(4-3)\cdot (1+3)^3}}=-\frac{5}{16}[/m]
[m]f(3,2)=\sqrt{\frac{4-3,2}{1+3,2}} [/m]
[b][m]\sqrt{\frac{4-3,2}{1+3,2}}≈ f(3) + f`(3)\cdot 0,2=\frac{1}{2}-\frac{5}{16}\cdot 0,2=[/m] [/b] считайте
2)
[m]f(x)=arctg\sqrt{x}[/m]
x_(o)=1
x_(o)+ Δx=0,97
Δx=-0,03
[m]f(1)=arctg\sqrt{1}=\frac{π}{4}=[/m]
[m]f`(x)=(arctg\sqrt{x})`=\frac{1}{1+(\sqrt{x})^2}\cdot (\frac{1}{2\sqrt{x}}[/m]
[m]f`(1)=\frac{1}{1+(\sqrt{1})^2}\cdot (\frac{1}{2\sqrt{1}})=0,25[/m]
[b][m]arctg\sqrt{0,97} ≈\frac{π}{4}+0,25\cdot (-0,03)= [/m][/b]считайте