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$$f(x)=\begin{cases} \frac{x^2-1}{x+1},&\text{if }x<-4\\\\ ax^2+2x+b,&\text{if }-1\le x<0\\\\ |x+a+2|,&\text{if }x\ge 0\;. \end{cases}$$

I usually can solve these kinds of questions but the fact that the function is not defined on $(-4,-1)$ throws me off.

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Ignore what happens around $-4$ and solve for $-1$ and $0$. The function $f$ can't be continuous where it isn't defined, namely on $(-4,-1)$. Note that it can't be discontinuous on that interval either. It simply doesn't make sense to talk about continuity outside the domain of $f$. –  Git Gud Feb 1 '13 at 22:32

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Note that $$\frac{x^2-1}{x+1}=x-1$$ if $x\ne-1$, so the definition of $f$ can be simplified to

$$f(x)=\begin{cases} x-1,&\text{if }x<-4\\ ax^2+2x+b,&\text{if }-1\le x<0\\ |x+a+2|,&\text{if }x\ge 0\;. \end{cases}$$

This function is certainly continuous on the open ray $(\leftarrow,-4)$. We don’t have to worry about continuity at any point of $[-4,-1)$, because those points aren’t in the domain of the function. Thus, the problem really boils down to choosing $a$ and $b$ so that the function

$$g(x)=\begin{cases} ax^2+2x+b,&\text{if }-1\le x<0\\ |x+a+2|,&\text{if }x\ge 0 \end{cases}$$

is continuous on the closed ray $[-1,\to)$. That’s not hard to do, but there are infinitely many pairs $\langle a,b\rangle$ that work. Consequently, I strongly suspect that there may be a misprint in the problem, and that the first case of the definition of $f$ was supposed to be $x<-1$.

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I suspected that as well, but I checked and it seems it was a misprint but then they decided to maintain the question as it is. So the answers for a and b will be defined by each other and be all real numbers, correct? –  Kev Feb 1 '13 at 22:39
    
@Kev: Not quite: you have to have $b=|a+2|$, so $b$ has to be non-negative. For $b=0$ the only usable value of $a$ is $-2$, and for each $b>0$ there will be two usable values of $a$. –  Brian M. Scott Feb 1 '13 at 22:41
    
Can't I remove the absolute value brackets to make it a+2? When the function approaches zero from the right, x is always positive. It should work for f(0) as well. –  Kev Feb 1 '13 at 22:49
    
@Kev: No, you can’t. It isn’t important whether $x$ is positive, because you’re not taking the abs. value of $x$: you’re taking the abs. value of $x+a+2$, which certainly need not be positive just because $x$ is. –  Brian M. Scott Feb 1 '13 at 22:50
    
Alright. If the condition a>0 was added into the question would I be able to remove the absolute value brackets? As I understand it if you know for a fact that the value in the bracket is always positive, you can simply remove the brackets. –  Kev Feb 1 '13 at 22:53

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