数学问题:求函数y=sinx+c
1,求函数y=sinx+cosx+sinxcosx的值域
令:t=sinx+cosx=√2sin(x+π/4)--->-√2≤t≤√2
--->t²=(sin²x+cos²x)+2sinxcosx--->sinxcosx=(t²-1)/2
--->y = sinx+cosx+sinxcosx
= t+(t²-1)/2 = (1/2)(t+1)²-1
--->t=-1时,y有最小值1;
t=√2时,y有最大值√2+(1/2)
--->y的值域为[-1,√2+(1/2)]
2,求函数y=(x+1)(x+2)(x-3)(x-4...全部
1,求函数y=sinx+cosx+sinxcosx的值域
令:t=sinx+cosx=√2sin(x+π/4)--->-√2≤t≤√2
--->t²=(sin²x+cos²x)+2sinxcosx--->sinxcosx=(t²-1)/2
--->y = sinx+cosx+sinxcosx
= t+(t²-1)/2 = (1/2)(t+1)²-1
--->t=-1时,y有最小值1;
t=√2时,y有最大值√2+(1/2)
--->y的值域为[-1,√2+(1/2)]
2,求函数y=(x+1)(x+2)(x-3)(x-4)+6的值域
y = [(x+1)(x-3)][(x+2)(x-4)] + 6
= (x²-2x-3)(x²-2x-8) + 6
= [(x-1)²-4][(x-1)²-9] + 6
= (x-1)^4-13(x-1)²+42
= [(x-1)²-13/2]²-1/4
≥ -1/4
--->值域为[-1/4,+∞)
3,已知x+2y+3z=12,求证:x²+2y²+3z²≥24
由柯西不等式:(x²+2y²+3z²)(1+2+3)≥(x+2y+3z)²=144
---> x²+2y²+3z²≥24
4,已知1≤x²+y²≤2,求x²-xy+y²的取值范围
设:x=rcost,y=rsint,1≤r²≤2
--->x²-xy+y²=r²-(r²/2)sin2t=(r²/2)(2-sin2t)
--->sin2t=-1,r²=2时,x²-xy+y²有最大值3
sin2t=1,r²=1时,x²-xy+y²有最大值1/2
--->1/2≤x²-xy+y²≤3。
收起