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# X is bounded thus the derivative at x = 0 exists and

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Unformatted text preview: x is bounded. Thus the derivative at x = 0 exists and is equal to 0. 2 3. (25 pts) (a) (15 pts) Suppose that f is differentiable on a nonempty, open interval ( a, b ), with f bounded on ( a, b ). Prove that f is uniformly continuous on ( a, b ). Solution: By assumption ∃ M > 0 such that | f ( x ) | ≤ M , for any x ∈ ( a, b ). Let > 0 and take δ = /M . Let y, z ∈ ( a, b ) such that | z- y | < δ . By the mean value theorem applied to f on the interval between y and z , there exists c such that f ( z )- f ( y ) = f ( c )( z- y ). Thus | f ( z )- f ( y ) | = | f ( c ) || z- y | ≤ M | z- y | < Mδ = . Since this is true for any y, z such that | z- y | < δ , it follows that f is uniformly continuous on ( a, b ). 2 (b) (10 pts) Give an example to show that if the hypothesis f bounded on ( a, b ) is omitted, then the statement of (a) is no longer true. Solution: Let f : (0 , 1) → R , defined by f ( x ) = ln x . f is differentiable on (0 , 1), and its derivative f ( x ) = 1 /x is clearly not bounded on (0 , 1). The function f is not uniformly continuous because it cannot be extended by continuity at 0 (the limit of f as x → + is-∞ ). 2 4. (15 pts) Show that ln( x + 1) ≤ x , for all x ≥ 0....
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x is bounded Thus the derivative at x = 0 exists and is...

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