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Math Help - Claim this function is contin and diff

  1. #1
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    Claim this function is contin and diff

    Suppose that  f(x)=\left\{\begin{array}{cc}x^2,&\mbox{ if }<br />
x \ is \  irrational \\0, & \mbox{ if } x \in \mathbb {Q} \end{array}\right.

    Is f continuous at 0 and is f differentiable at 0?

    Claim: f is continuous at 0.

    Given  \epsilon > 0 , pick  \delta = \sqrt { \epsilon } , then whenever  |x-0|=|x|< \delta

    Now, |f(x)-f(0)|=|f(x)| = \left\{\begin{array}{cc}|x^2|,&\mbox{ if }<br />
x \ is \  irrational \\0, & \mbox{ if } x \in \mathbb {Q} \end{array}\right. < x^2 = \delta ^2 = \epsilon

    Q.E.D.

    Claim: f is differentiable.

    Proof.

    Now, if x_0 \in \mathbb {Q} , then  \lim _{h \rightarrow 0 } \frac { f(x_0+h)-f(0)}{h}= \lim _{h \rightarrow 0 } \frac {f(h)}{h}=0

    If x_0 is irrational, then  \lim _{h \rightarrow 0 } \frac { f(x_0+h)-f(0)}{h}= \lim _{h \rightarrow 0 } \frac { (x_0+h)^2-0}{h}= \lim _{h \rightarrow 0 } \frac {2x_0^2+2x_0h+h^2}{h} = 0

    So f is diff.

    Q.E.D.

    Is this right? Thanks.
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  2. #2
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    Quote Originally Posted by tttcomrader View Post
    Suppose that  f(x)=\left\{\begin{array}{cc}x^2,&\mbox{ if }<br />
x \ is \ irrational \\0, & \mbox{ if } x \in \mathbb {Q} \end{array}\right.

    Is f continuous at 0 and is f differentiable at 0?

    Claim: f is continuous at 0.

    Given  \epsilon > 0 , pick  \delta = \sqrt { \epsilon } , then whenever  |x-0|=|x|< \delta

    Now, |f(x)-f(0)|=|f(x)| = \left\{\begin{array}{cc}|x^2|,&\mbox{ if }<br />
x \ is \ irrational \\0, & \mbox{ if } x \in \mathbb {Q} \end{array}\right. < x^2 = \delta ^2 = \epsilon

    Q.E.D.

    Claim: f is differentiable.

    Proof.

    Now, if x_0 \in \mathbb {Q} , then  \lim _{h \rightarrow 0 } \frac { f(x_0+h)-f(0)}{h}= \lim _{h \rightarrow 0 } \frac {f(h)}{h}=0

    If x_0 is irrational, then  \lim _{h \rightarrow 0 } \frac { f(x_0+h)-f(0)}{h}= \lim _{h \rightarrow 0 } \frac { (x_0+h)^2-0}{h}= \lim _{h \rightarrow 0 } \frac {2x_0^2+2x_0h+h^2}{h} = 0

    So f is diff.

    Q.E.D.

    Is this right? Thanks.
    On the differentiable part you are showing that the derivative exits at 0 so x_0=0 and it is not irrational

    You can do this with only one case, but you have the right idea

    The defintion you are using is this

    f'(c)=\lim_{h \to 0}\frac{f(c+h)-f(c)}{h}=\lim_{h \to 0}\frac{f(0+h)-f(0)}{h}=\lim_{h \to 0}\frac{f(h)-f(0)}{h}

    You may want to use the squeeze thorem

    Note that 0 \le f(x) \le x^2 for all x.


    Good luck.
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  3. #3
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     \lim _{h \rightarrow 0 } \frac {[f(h)]^2}{h} \geq \lim _{h \rightarrow 0 } \frac {f(h)}{h} \geq 0

    Is it something I can do with the left hand side? Does it involve rationality?
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  4. #4
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    Quote Originally Posted by tttcomrader View Post
     \lim _{h \rightarrow 0 } \frac {[f(h)]^2}{h} \geq \lim _{h \rightarrow 0 } \frac {f(h)}{h} \geq 0

    Is it something I can do with the left hand side? Does it involve rationality?

    Question 1:

    Is this statement true:

    0 \le f(x) \le x^2 for ALL(both rational and irrational) values of x.

    If the above is true, how does this help you?

    0 \le \lim_{h \to 0}\frac{f(0+h)-f(0)}{h} \le \lim_{h \to 0}\frac{(0+h)^2-0^2}{h}=\lim_{h \to 0}h
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