tìm max min của hs

N

nguyenbahiep1

a)

[TEX]y = 2x + \sqrt{5-x^2} \\ txd:- \sqrt{5} \leq x \leq \sqrt{5} \\ y ' = \frac{2.\sqrt{5-x^2}-x}{\sqrt{5-x^2}}= 0 \\ \Rightarrow x= 2 \\ x= -2 \\ min f(-\sqrt{5}) = -2.\sqrt{5} \\ max f(2) = 5[/TEX]

b)

[TEX]y = \frac{3.x^2+3}{2x^2+x+2} \\ txd: x : R \\ y' = \frac{3x^2-3}{(2x^2+x+2)^2} = 0 \\ \Rightarrow x = -1 \\ x= 1 \\ \Rightarrow min f(1) = \frac{6}{5} \\ max f(-1) = 2[/TEX]

c)

[TEX]y= \frac{x^4 + 2}{x^2} \\ txd : x \not= 0 \\ y' = \frac{2x^4-4}{x^3} = 0 \\ x = \sqrt[4]{2} \\ x = -\sqrt[4]{2} \\ \Rightarrow min f( \sqrt[4]{2}) = f( -\sqrt[4]{2}) =2.\sqrt{2}[/TEX]
 
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