Check whether the function, ,f defined below, is uniformly continuous or not:
f(x)=x^(1/2), x∈[1,2]
1
Expert's answer
2022-03-15T13:11:09-0400
ANSWER: The function f(x)=x21 is uniformly continuous on [1,2]
EXPLANATION 1.
Let ϵ>0 . We need to show that exists δ>0 such that
x,y∈[1,2] and ∣x−y∣<δ imply ∣f(x)−f(y)∣<ϵ .
We have
f(x)−f(y)=x−y=x+y(x−y)⋅(x+y)=x+yx−y .
Because x≥1,y≥1 , then x+y∣x−y∣≤2∣x−y∣ . If δ=2ϵ and ∣x−y∣≤δ , then ∣f(x)−f(y)∣≤2∣x−y∣≤2δ=22ϵ=ϵ . So, we have shown, by the definition, that f is uniformly continuous on [1,2].
EXPLANATION 2.
The function f is continuous on a closed interval [1,2] , so by Cantor's theorem the function f is uniformly continuous on [1,2].
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