Appl of IVT - ≤ 0 ≤ f (0) – 0 = g (0) (as 0 ≤ f ( x...

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Problem Proof. Let g : [0, 1] R be the function defined by g ( x ) = f ( x ) x . Note that (i) g is continuous (as f is continuous) on [0, 1] and (ii) g (1) = f (1) – 1
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Unformatted text preview: ≤ 0 ≤ f (0) – 0 = g (0) (as 0 ≤ f ( x ) ≤ 1). Thus, by IVT , there exists c in [0, 1] s.t. g ( c ) = 0, i.e., f ( c ) = c ....
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