MHB How can I ensure continuity for a piecewise function with a radical term?

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To ensure continuity for a piecewise function involving a radical term, it is essential to recognize that the cube root function, \(x^{1/3}\), is continuous for all real numbers, including negative values. The confusion arises with even roots, such as \(\sqrt[n]{x}\), which are not defined for negative inputs. For a function to be continuous at a specific point, the left-hand limit, right-hand limit, and the function's value at that point must all be equal. The discussion clarifies that the constants a and b can be chosen to maintain continuity, provided the correct radical function is used. Understanding the properties of radical functions is crucial for determining continuity across the entire real line.
Petrus
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Hello MHB,
If I want to decide constant a and b so its continuous over the whole R for this piecewise function
102oyec.png

basicly what I got problem with is that $$x^{1/3}$$ is not continuous for negative value so it will never be continuous for any value on constant a,b. I am missing something? or do they mean $$\frac{1}{8}$$ and not $$-\frac{1}{8}$$

Regards,
$$|\pi\rangle$$
 
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Petrus said:
Hello MHB,
If I want to decide constant a and b so its continuous over the whole R for this piecewise function
102oyec.png

basicly what I got problem with is that $$x^{1/3}$$ is not continuous for negative value so it will never be continuous for any value on constant a,b. I am missing something? or do they mean $$\frac{1}{8}$$ and not $$-\frac{1}{8}$$

Regards,
$$|\pi\rangle$$
In fact
\sqrt[3]{x} is defined for all real numbers but the problem is in
\sqrt[n]{x} with n even number
for a function to be continuous at a point c
\lim_{x \rightarrow c^- } f(x) = \lim_{x\rightarrow c^+ } f(x) = f(c)
 
Amer said:
In fact
\sqrt[3]{x} is defined for all real numbers but the problem is in
\sqrt[n]{x} with n even number
for a function to be continuous at a point c
\lim_{x \rightarrow c^- } f(x) = \lim_{x\rightarrow c^+ } f(x) = f(c)
Thanks for the fast respond you are totally correct! I confused myself! Have a nice day!

Regards,
$$|\pi\rangle$$
 
There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...

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