Question

已知fx=3-log2x,x属于[1,16]求y=[f(x)]2+f(x)2的值域

Answer 1

f(x)=3-log2x,x∈[1,16]

y=[f(x)]²+f(x²) = {3-log2[x]}² + 3 - log2[x²]

= 9 - 6log2[x] + {log2[x]}² + 3 - 2log2[x]

= {log2[x]}²-8log2[x]+12

= {log2[x]-4}² - 4

x∈[1,16]

log2[x]∈[0,4]

log2[x]-4∈[-4,0]

{log2[x]-4}² ∈[0,16]

{log2[x]-4}² ∈[-4,12]

值域：[-4,12]