求f(x)=4的x方-2的（x+2）方+5,x∈[1,3]的值域.

求f(x)=4的x方-2的（x+2）方+5,x∈[1,3]的值域.
求f(x)=4的x方-2的（x+2）方+5,x∈[1,3]的值域.

求f(x)=4的x方-2的（x+2）方+5,x∈[1,3]的值域.
f(x)=(2^x)^2-2^x*4+5=(2^x-2)^2+1
又1<=x<=3,则2<=2^x<=8
0<=2^x-2<=6.
0<=(2^x-2)^2<=36
所以,1<=f(x)<=37
即值域是[1,37]