✔ 最佳答案
極值發生於一次微分等於零處
dy/dx = { (2X) ' (X^2 +1) -2X(X^2 +1) ']/ (X^2+1)^2
={2 (X^2 +1) -2X(2X)) } /(X^2+1)^2
=( -2X^2 +2 ) / (X^2+1)^2
當X =+1 ,-1 時 dy/dx=0
x=1 , y= 2/ (1+1) =1
x=-1 , y= -2 /(1+1) =-1
Ans :最大值1 , 最小值-1
2014-11-17 20:50:30 補充:
y=(2x)/(x^2+1) ={ (x^+2x+1) - (x^2 +1) } / (x^2+ 1) = (x+1)^2 /(x^2+1) -1
∵ (x+1)^2 /(x^2+1) ≧0 ∴ y≧ 0-1 = -1
=> y最小值 -1
∵ (x-1)^2 ≧0 => x^2 -2x +1≧0 =>x ^2+1 ≧2x
∴ y=(2X)/(X^2+1) ≦ (x^2 +1 )/ (x^2 +1) =1
=> y最大值1
2014-11-17 20:53:36 補充:
補充漏字更正
y=(2x)/(x^2+1) ={ (x^2 +2x+1) - (x^2 +1) } / (x^2 + 1) ={ (x+1)^2 } /(x^2 +1) -1
∵ (x+1)^2 /(x^2+1) ≧0 ∴ y≧ 0-1 = -1
=> y最小值 -1
∵ (x-1)^2 ≧0 => x^2 -2x +1≧0 =>x ^2 +1 ≧2x
∴ y=(2x)/(x^2+1) ≦ (x^2 +1 )/ (x^2 +1) =1
=> y最大值1