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Fix $\alpha ,\beta\in\mathbb R$ and define a function $f$ from $\mathbb R$ to $\mathbb R$ such that $$f(x)= \begin{cases} x^{\alpha} + 3 & \quad \text{if } x\le 1 \\ x^{\beta} + 3 & \quad \text{otherwise} \end{cases}$$

How do I prove that f is continuous?

Any hint? I am stuck.

Thank you

José Carlos Santos
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xNJL
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  • Your function is not well defined for some values of $\alpha, \beta \in \mathbb{R}$ –  Dec 08 '17 at 12:33
  • @Math_QED How is it not well defined? – Toby Mak Dec 08 '17 at 12:34
  • For example, what is $(-1)^{1/2}$? –  Dec 08 '17 at 12:36
  • Even if the point doesn't exist, all that matters is that $1)$ - $f(1)$ exists, $2)$ - the limit of $x$ to $1$ of $f(x)$ exists, and that $3)$ - $1)$ is equal to $2)$. This post proves that we can express $a^b$ as $e^{b \ln a}$, so $1^b$ is well-defined for real numbers. – Toby Mak Dec 08 '17 at 12:39
  • Well, then what is $\ln(-1)$ in the real numbers? Clearly this function is badly defined for certain values as the output does not give real numbers. –  Dec 08 '17 at 12:40
  • @Math_QED $x$ can't be changed to $-1$. – Toby Mak Dec 08 '17 at 12:43
  • But the function is defined on the real numbers, in particular on the negative real numbers? –  Dec 08 '17 at 12:44
  • You are right, we should bound alpha and beta to obtain a well defined function – xNJL Dec 10 '17 at 11:26

1 Answers1

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It's just a matter of noticing that$$\lim_{x\to1^-}f(x)=\lim_{x\to1^+}f(x)=4=f(1).$$

José Carlos Santos
  • 427,504