Pls help me do Calc corrections!

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pluralistic

Pls help me do Calc corrections!

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(Based on todays review)

LammettHash

1. $a=2,b=3$ makes it so that

$\displaystyle\lim_{x\to1^-}f(x)=\lim_{x\to1}(3-x)=2$
$\displaystyle\lim_{x\to1^+}f(x)=\lim_{x\to1}(2x^2+3x)=5$

and because the one-sided limits do not agree, $f(x)$ is not continuous at $x=1$ under these conditions.

2. For $f(x)$ to be continuous, we need both limits to match up; specifically, we require

$\displaystyle\lim_{x\to1}(ax^2+bx)=a+b=2$

3. Similar to (2), we need the limits from either side of $x=2$ to agree:

$\displaystyle\lim_{x\to2^+}f(x)=\lim_{x\to2}(5x-10)=0$
$\implies\displaystyle\lim_{x\to2^-}f(x)=\lim_{x\to2}(ax^2+bx)=4a+2b=0\implies 2a+b=0$

4. $f(x)$ will be continuous everywhere so long as $a,b$ are chosen such that both $a+b=2$ and $2a+b=0$ .

$a+b=2\implies b=2-a\implies 2a+(2-a)=a+2=0\implies a=-2\implies b=4$

5. I'll leave this part to you.
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