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Let $a,b,c$ be real numbers defined such that:

$f(x) = \begin{cases} \displaystyle\frac{1}{|x|}, & \mbox{if } |x| >c \\ \\ a+bx^2, & \mbox{if } |x| \leq c \end{cases}$

Find $a,b$ in terms of $c$ such that $f$ is differentiable at $c$.

The issue I am having is what cases do I have to consider? I think $c<0$ and $c\geq 0$ . But then, what is the expression of $f(x)$ in each case ?

Thank you in advance

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    I do $n$ot k$n$ow.. that is my question2012-10-14

2 Answers 2

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What would it mean if we write $|x|>c$ and $|x|, when $c$ be a negative real number, in the rule of $f(x)$ simultaneously? $c$ should be a positive real number. If so, then your function will be: $f(x) = \begin{cases} \displaystyle\frac{1}{x}, & \mbox{if } x >c \\ \displaystyle\frac{1}{-x}, & \mbox{if } x < -c \\ a+bx^2, & \mbox{if } |x| \leq c \end{cases}$ I think, now, you can use the method @Jean suggested above.

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    :-) ${}{}{}{}{}{}$, @amWhy2013-03-28
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Hint: Consider the case $c=0$ separatly, and then use the definition (left limit, right limit) of the derivative to find $a$ and $b$ for all values of $c$

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Show that $ f(x) = \begin{cases} \displaystyle\frac{1}{x}, & \mbox{if } x >c \\ \\ a+bx^2, & \mbox{if } -c\leq x \leq c\\ \frac{-1}{x} &\mbox{if } x <-c\end{cases} $ with $c\geq 0.$

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    I have one more question. In order to show that we do not have to consider the case c<0, what should I write? Is it enough to say that since $\mid x \mid \geq 0$, and since either &\mid$x$\mid >c& or $\mid x \mid \leq c$ then $c \geq 0$ ?2012-10-14