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Math Help - Piecewise function with Taylor's Theorem

  1. #1
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    Piecewise function with Taylor's Theorem

    Let U be a neighborhood of 0 in  \mathbb {R} .

    Suppose that f:U \rightarrow \mathbb {R} is twice differentiable , f' and f'' are continuous, and f(0)=0.

    Define g: U \rightarrow \mathbb {R} by g(x)= \frac {f(x)}{x} if x \neq 0 and  \lim _{ h \rightarrow 0 } \frac {f(h)}{h} if  x = 0

    Show that g(0) exists and that g' exist and is continuous on U. Find g(0),g'(0)

    Proof so far.

    Find g(0):

    By the Taylor's Theorem, I can write f(x)=f(0)+f'(0)x+f''(t)x^2=f'(0)x+f''(t)x^2

    Now, g(0)= \lim _{h \rightarrow 0 } \frac {f(h)}{h} = \lim _{h \rightarrow 0 } \frac {f(h)-f(0)}{h} = f'(0)

    Is that right so far?

    And how do I find g'(0)? Thank you.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by tttcomrader View Post
    Let U be a neighborhood of 0 in  \mathbb {R} .

    Suppose that f:U \rightarrow \mathbb {R} is twice differentiable , f' and f'' are continuous, and f(0)=0.

    Define g: U \rightarrow \mathbb {R} by g(x)= \frac {f(x)}{x} if x \neq 0 and  \lim _{ h \rightarrow 0 } \frac {f(h)}{h} if  x = 0

    Show that g(0) exists and that g' exist and is continuous on U. Find g(0),g'(0)

    Proof so far.

    Find g(0):

    By the Taylor's Theorem, I can write f(x)=f(0)+f'(0)x+f''(t)x^2=f'(0)x+f''(t)x^2
    No you can't, there has to be a remainder term.

    Now, g(0)= \lim _{h \rightarrow 0 } \frac {f(h)}{h} = \lim _{h \rightarrow 0 } \frac {f(h)-f(0)}{h} = f'(0)

    Is that right so far?
    Since f(0)=0:

     <br />
g(0)=\lim_{h \to 0} \frac{f(h)}{h}=\lim_{h \to 0} \frac{f(h)-f(0)}{h}=f'(0)<br />

    CB
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  3. #3
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    Do I find g'(0) the same way then?

    g'(0) = \lim _{h \rightarrow 0 } \frac {g(h)-g(0)}{h} = \lim _{h \rightarrow 0 } \frac { \frac {f(h)}{h}-f'(0)}{h} = \frac {g(0)-g(0)}{h} = 0

    Is this right?
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