Given f : R → R is continuous on [a,b] and sgn( f(a) )sgn( f(b) )= -1
Show that ∃ x∈[a,b] : f(x) = 0
Looks to me like a variation of Rolle's theorem.
No, I would restate it as "Many [relatively] easy statements in mathematics are [relatively] hard to prove."
Two cases: either
f(a) < 0, f(b) > 0
or
f(a) > 0, f(b) < 0
Given f(x) is continuous from a to b, it must not jump. And yet it must cross 0 in order to go from <0 to >0 (or the other way around). Therefore it must at some point equal 0.
Is this proof valid?
I would like to get a more Rigour proof
i'm not the best at writing proofs, sorry
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Looks to me like a variation of Rolle's theorem.
No, I would restate it as "Many [relatively] easy statements in mathematics are [relatively] hard to prove."
Two cases: either
f(a) < 0, f(b) > 0
or
f(a) > 0, f(b) < 0
Given f(x) is continuous from a to b, it must not jump. And yet it must cross 0 in order to go from <0 to >0 (or the other way around). Therefore it must at some point equal 0.
Is this proof valid?
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