0 < X < 1.
The above inequality states that x can be any value BETWEEN 0 and 1,but 0 and 1 are NOT included.
F(1/2) =(1 + 1/2) (1 + (1/2)^4)(1 + (1/2)^16)(1 + (1/2)^64)(1 + (1/2)^256),
F(1/2) = (3/2)(17/16)(1 + 1/2^16)(1 + 1/2^64)(1 + 1/2^256),
F(1/2) =(3/2)(17/16)(1)(1)(1)= 1.59375.
For 0 < x < 1, let
f(x) = (1 + x)(1 + x4)(1 + x16)(1 + x64)(1 + x256) · · ·
Compute f−1
8
5f(3/8)
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