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解:
0<=x<=2
y=4x-1/2-3*2^x+5
设y=f(x)=4x-1/2-3*2^x+5
y'=f(x)'=(4x-1/2-3*2^x+5)'=4-3*2^x ln2
令y'=f(x)'=4-3*2^x ln2>=0
解得:2^x<=4/3ln2
x<=log2(4/3ln2)
所以当0<=x<=log2(4/3ln2)时,f(x)单调递增
在log2(4/3ln2)<=x<2时, f(x)单调递减
即当x=log2(4/3ln2)时ymax=f(log2(4/3ln2))=4*log2(4/3ln2)-1/2-3*2^(log2(4/3ln2))+5
y0=f(0)=4*0-1/2-3*2^0+5=3/2
y2=f(2)=4*2-1/2-3*2^2+5=1/2
因为y2<y0
所以ymin=y2=1/2
欢迎采纳!
0<=x<=2
y=4x-1/2-3*2^x+5
设y=f(x)=4x-1/2-3*2^x+5
y'=f(x)'=(4x-1/2-3*2^x+5)'=4-3*2^x ln2
令y'=f(x)'=4-3*2^x ln2>=0
解得:2^x<=4/3ln2
x<=log2(4/3ln2)
所以当0<=x<=log2(4/3ln2)时,f(x)单调递增
在log2(4/3ln2)<=x<2时, f(x)单调递减
即当x=log2(4/3ln2)时ymax=f(log2(4/3ln2))=4*log2(4/3ln2)-1/2-3*2^(log2(4/3ln2))+5
y0=f(0)=4*0-1/2-3*2^0+5=3/2
y2=f(2)=4*2-1/2-3*2^2+5=1/2
因为y2<y0
所以ymin=y2=1/2
欢迎采纳!
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