Continuity with the following function

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SUMMARY

The function h : ℝ → ℝ is defined piecewise, with h(x) = 0 for x in ℚ and h(x) = x³ + 3x² for x not in ℚ. It is continuous at x = 0 and x = -3, while being discontinuous at all other points. The proof involves evaluating limits from rational and irrational sequences. For differentiability, h is differentiable at x = 0, but non-differentiable elsewhere, which can be shown using the limit definition of the derivative and sequences.

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Caeder
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Define [tex]h : \mathbb{R} \rightarrow \mathbb{R}[/tex]

[tex]h(x) = \begin{cases} 0 &\text{if\ }\ x \in \mathbb{Q}\\ x^3 + 3x^2 &\text{if\ }\ x \notin \mathbb{Q} \end{cases}[/tex].



a.) Determine at what points [tex]h[/tex] is continuous and discontinuous. Prove results.

b.) Determine at what points [tex]h[/tex] is differentiable and non-diff'able. Prove results.

My work:

[tex]h[/tex] is obviously cont. when [tex]0=x^3 + 3x^2[/tex] as when approaching x, lim is 0 by rationals and lim is [tex]x^3 + 3x^2[/tex] by irrationals. So, it's cont. at [tex]x=0,-3[/tex] and discont. everywhere else.

Not sure how to prove this.

And for differentiability.. not sure.. diff'able nowhere?
 
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maybe diffble at 0?
 
use sequences to prove (a), remember the theorem that f is continuous on a set E if for every sequence x_n in E that converges to x_0 in E has the property that f(x_n) converges to f(x_0).

for (b) u can use sequences to find out where lim[(f(x)-f(y))/(x-y)] as x goes to y, where y is fixed exists.
 

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