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# 10.28 - Section 3.2 Inverse functions and...

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Section 3.2: Inverse functions and logarithms (concluded) Fact: If f is an increasing function then f is one-to-one. (Note: Here “increasing” means “strictly increasing”.) Proof: If x 1 , x 2 are in the domain of f with x 1 < x 2 , then since f is increasing, f ( x 1 ) < f ( x 2 ), so that f ( x 1 ) f ( x 2 ). Likewise, every decreasing function is one-to-one. These results have a kind of converse: Theorem: If f is a continuous function whose domain is some interval I , and f is one-to-one, then f is either increasing on all of I or decreasing on all of I . In either case, the inverse function f –1 is also continuous. (We’ll prove this later using the Mean Value Theorem.) What about differentiability? Theorem 7: Let f be a one-to-one differentiable function, and let a = f –1 ( b ) for some number b in the range of f . If f ( a ) 0, then f –1 is differentiable at b , and ( f –1 ) ( b ) = 1 / f ( a ).

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Example: Let f ( x ) = x 3 , b = 8, and a = f –1 (8) = 2. f (2) = 3 (2) 2 = 12 0, so ( f –1 ) (8) = 1/12.
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