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How to solve this question:

Compute the value of A that will make the function f(x) continuous everywhere.

$$f(x)=\begin{cases}x^2-2&,\;\;\text{if}\;\;x=-1\\Ax-4&,\;\;\text{if}\;\;x\neq -1\end{cases}$$

Is it necessary solve this answer with graphs??

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What is the value of $f(-1)$? (Some work is required!) –  copper.hat Jan 8 '13 at 5:12

4 Answers 4

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Note that $$\lim_{x \to -1} f(x) = \lim_{x \to -1} (Ax-4) = -A-4$$ $$f(-1) = (-1)^2 - 2 = -1$$ For $f(x)$ to be continuous at $x=-1$, we need $$\lim_{x \to -1} f(x) = f(-1)$$ Hence, we get that $$-A - 4 = -1 \implies A =-3$$

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You have to show that $$\lim_{x\rightarrow-1^{-}}{f(x)}=\lim_{x\rightarrow-1^{+}}{f(x)}=f(-1)$$

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Bottom line, $f(-1)=(-1)^2-2=-1\Longrightarrow$ the value of the limit when $\,x\to -1\,$ must equal this, so:

$$\lim_{x\to -1}(Ax-4)=-A-4=-1\Longleftrightarrow A=-3$$

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For limit Continuous we can assume the X as -1 where initially when x=-1 whe has to use first function ie f(x)= x*x-2 and use it in second function that is Ax-4

A=-3

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