求y=sqr(2x+2)+sqr(1-x)的值域RT
问题描述:
求y=sqr(2x+2)+sqr(1-x)的值域
RT
答
b=√(2x+2)c=√(1-xb²/4=x+1c²=1-xb²/8+c²/2=1b>=0,c>=0则b=√8cosac²/2=1-cos²a=sin²ac=√2sinaa是锐角y=b+c=√2sina+√8cosa=√10sin(a+d)其中tand=√8/√2=2>1所以45...