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Problem:

Let $f(x) = \left\{ \begin{array}{lr} \frac{\arctan(x)}{(1+x)^2} & : x \geq 0\\ Ae^x + B & : x < 0 \end{array} \right. $

Find $A$ and $B$ such that the function is continuous and differentiable at $x=0$.

My attempt:

To ensure continuity at $x = 0$ I figured $A = B = 0$ would be the only option. But this, of course, seems very wrong, and in any case, it wouldn't cause differentiability at $x=0$.

As far as I could tell, the derivative of $\frac{\arctan(x)}{(1+x)^2}$ at $x=0$ would be $1$. While the second piece's derivative evaluates trivially to $0$.

Git Gud
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Alec
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    "To ensure continuity at $x = 0$ I figured $A = B = 0$ would be the only option". Once you fix this mistake, you should be able to get the correct answer. What is $\lim \limits_{x\to 0^{\mathbf -}}(f(x))$? – Git Gud Oct 21 '14 at 09:40

2 Answers2

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$f$ is continuous at $0$ means that \begin{align*} \lim_{x\to0^+}f(x)&=\lim_{x\to0^-}f(x) \\\iff0&=A+B. \end{align*} Also, $f$ is differentiable at $0$ means that \begin{align*} \lim_{x\to0^+}f'(x)&=\lim_{x\to0^-}f'(x) \\\iff1&=A. \end{align*} (actually, this is valid since each piecewise defined part is smooth in its given domain.) Thus we have $A=1,B=-1$.

Jaehyeon Seo
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  • @GitGud Oh, my mistake. I edited. – Jaehyeon Seo Oct 21 '14 at 10:17
  • Also $f$ being differentiable at $0$ doesn't mean that $\lim \limits_{x\to0^+}f'(x)=\lim \limits_{x\to0^-}f'(x)$. In fact $\lim \limits_{x\to0^+}f'(x)=\lim \limits_{x\to0^-}f'(x)$ means that $f$ is differentiable and $f'$ is continuous and this is well known not to be necessarily true. – Git Gud Oct 21 '14 at 10:22
  • @GitGud To be precisely, yes. What we actually have to do is using the definition of derivitive and check that whether the limit exists. But I think "thinking in this way" is reasonable since both piecewise defined funtion are smooth in their given domain. We can prove that $\lim_{x\to a^+}g'(x)=\lim_{x\to a^+}(g(x)-g(a))/(x-a)$ if $g$ is smooth and so on. – Jaehyeon Seo Oct 21 '14 at 10:36
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Hint: $f$ is differentiable in 0 iff you have in a neighborhood of 0:

$$ f(x) = a + bx + o(x) $$

Here when $x>0$ you have $$ f(x) = x + o(x) $$so it is mandatory that $$a=0, b=1 $$

but when $x<0$ you have $$f(x) = A(1+x + o(x)) + B $$

mookid
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