- #1
Caeder
- 13
- 0
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?
[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?