如果题目是:在曲线y=x/(1+x^2)上求一点,使通过该点的切线平行于X轴,并求出此切线的方程
y = x/(1+x²)
y ' =[ (1+x²) - 2x²]/(1+x²)² =(1-x²)/(1+x²)² = 0
解得:x=1或 x = -1
把x =1代入y = x/(1+x²)得:y = 1/2
把x =-1代入y = x/(1+x²)得:y = -1/2
所以过点(1,1/2) 的切线方程:y = 1/2
过点(1,-1/2)的切线方程是:y = -1/2
如果题目是:在曲线y=x/(1+x^2)上求一点,使通过该点的切线平行于X轴,并求出此切线的方程
y = x/(1+x²)
y ' =[ (1+x²) - 2x²]/(1+x²)² =(1-x²)/(1+x²)² = 0
解得:x=1或 x = -1
把x =1代入y = x/(1+x²)得:y = 1/2
把x =-1代入y = x/(1+x²)得:y = -1/2
所以过点(1,1/2) 的切线方程:y = 1/2
过点(1,-1/2)的切线方程是:y = -1/2