Determination of a constant based on continuity

Question

The following defines function with a constant $b$ to be determined by using the continuity of a function: $$f(x)=\begin{cases} \dfrac{x-b}{b+1}, \quad x<0\\[1.75ex] x^2+b, \quad x>0 \end{cases}$$

In short, for what value of $b$ is $f(x)$ continuous for every $x$.

What I did was to determine the right hand and left hand limit at zero to find that $b=0$ or $b=-2$ but I find this weird.

Ohhhhhhhhhh I think I got it @dfeuer, is it because the domain is defined as $(-\infty,0)\cup(0,\infty)$ — azetina, Nov 10 '13 at 20:16

score 3 · Accepted Answer · answered Nov 10 '13 at 20:11

3

If we require continuity we have

$$\lim_{x\to0^+}f(x)=b=\lim_{x\to0^-}f(x)=\frac{-b}{b+1}$$ Can you solve this equation to find the value of $b$?

answered Nov 10 '13 at 20:11

Yeah I did and got $0$ and $-2$. – azetina Nov 10 '13 at 20:12
Yes so there's two solutions. – Nov 10 '13 at 20:13
So what happens at $x=0$? – azetina Nov 10 '13 at 20:14
At $x=0$ in the case of continuity we have $f(0)=\lim _{x\to0}f(x)$. – Nov 10 '13 at 20:16
But $0\notin \mathbb{D}_f$? – azetina Nov 10 '13 at 20:18
1

You should understand that the function $f$ is continuous on $\mathbb R^*$ for all $b$ but the question is to find $b$ such that the function $f$ is continuous at $0$ so it will be defined at $0$ and its value is $\lim_{x\to0}f(x)$. – Nov 10 '13 at 20:22

Determination of a constant based on continuity

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