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alyannahere

  • 3 years ago

Determine the values of a and b for which the following function is continuous at every real number x f(x) { ax+b if x ≤ 1 -2bx^2 - 8a if 1 < x < 2 4b^2/a -5x if x ≥ 2 }

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  1. sriramkumar
    • 3 years ago
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    continuous implies \[\lim_{x \rightarrow c}f(x)=f(c)\] for the first one, \[\lim_{x \rightarrow 1}f(x)=f(1)\] \[\lim_{x \rightarrow 1}(ax+b)=a+b\] and so on...

  2. SithsAndGiggles
    • 3 years ago
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    \[f(x)=\begin{cases}ax+b&\text{if }x\le1\\ -2bx^2-8a&\text{if }1<x<2\\ \frac{4b^2}{a-5x}&\text{if }x\ge2\end{cases}\] For f(x) to be continuous at x = 1 and x = 2, the following must be satisfied: \[\color{red}{\lim_{x\to1^-}f(x)=\lim_{x\to1^+}f(x)}\\ \color{blue}{\lim_{x\to2^-}f(x)=\lim_{x\to2^+}f(x)}\] \[RED:\lim_{x\to1^-}(ax+b)=\lim_{x\to1^+}(-2bx^2-8a)\\ BLUE: \lim_{x\to2^-}(-2bx^2-8a)=\lim_{x\to2^+}\frac{4b^2}{a-5x}\] Determining the limits should give you a system of equations (assuming I rewrote the function you provided).

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