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Math Help - Is this function differentiable at 0?

  1. #1
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    Is this function differentiable at 0?

    For a natural number  n \geq 2 , define  f(x) =\left\{\begin{array}{cc}0,&\mbox{ if }<br />
x\leq 0\\x^n, & \mbox{ if } x>0\end{array}\right.

    Prove that f is differentiable at x = 0.

    Here what I did so far:

     \lim _{x \rightarrow 0^- } \frac {f(x)-f(0)}{x-0} = \lim _{x \rightarrow 0^- } \frac {f(x)}{x} = 0 since x approach from the left.

     \lim _{x \rightarrow 0^+ } \frac {f(x)-f(0)}{x-0} = \lim _{x \rightarrow 0^+ } \frac {x^n}{x} = 0 since x is approaching to 0.

    Is that right?
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by tttcomrader View Post
    For a natural number  n \geq 2 , define  f(x) =\left\{\begin{array}{cc}0,&\mbox{ if }<br />
x\leq 0\\x^n, & \mbox{ if } x>0\end{array}\right.

    Prove that f is differentiable at x = 0.

    Here what I did so far:

     \lim _{x \rightarrow 0^- } \frac {f(x)-f(0)}{x-0} = \lim _{x \rightarrow 0^- } \frac {f(x)}{x} = 0 since x approach from the left.

     \lim _{x \rightarrow 0^+ } \frac {f(x)-f(0)}{x-0} = \lim _{x \rightarrow 0^+ } \frac {x^n}{x} = 0 since x is approaching to 0.

    Is that right?
    If f(x) is differentiable at x=c this implies that \lim_{x\to{c^-}}\frac{f(x)-f(c)}{x-c}=\lim_{x\to{c^+}}\frac{f(x)-f(c)}{x-c}\implies\lim_{x\to{c}}\frac{f(x)-f(c)}{x-c}\text{ exists}

    So having  f(x) =\left\{\begin{array}{cc}0,&\mbox{ if }<br />
x\leq 0\\x^n, & \mbox{ if } x>0\end{array}\right. we have that

    \lim_{x\to{0^-}}\frac{f(x)-f(0)}{x-0}=\lim_{x\to{0^-}}\frac{0-0}{x}=0

    and

    \lim_{x\to{0^+}}\frac{f(x)-f(0)}{x-0}=\lim_{x\to{0^+}}\frac{x^{n}-0}{x}=\lim_{x\to{0^+}}x^{n-1} and since n\geq{2} we have that \lim_{x\to{0^+}}x^{n-1}=0

    So \lim_{x\to{0^-}}\frac{f(x)-f(0)}{x-0}=\lim_{x\to{0^+}}\frac{f(x)-f(0)}{x-0}=0. Thus f(x) is differentiable at x=0
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