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if you guys would turn to page 117 in Spivak's Calculus, there is the proof for theorem 3. At the last line he stated that this last inequality ##|f(x)-f(a)|<f(a)## implies ##f(x)>0##. How can you check this fact?

Can we assume first that ##f(x)-f(a)<0## to eliminate the absolute value, which leads to the inequality ##f(x)>0##?

On the second case if ##f(x)-f(a)>0##, we have ##2f(a)-f(x)>0##, how can we infer that ##f(x)>0##

Is it sufficient to show that ##f(x)>0## with the first case? (Logically this means ##(f(x)>0)\wedge (f(x)<2f(a))##

Thank You

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# The inequality which implies f(x) > 0 - Spivak's Calculus

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