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anonymous

  • 5 years ago

Check if the function is continuous.

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  1. anonymous
    • 5 years ago
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    \[f(x) = \left\{ {{{x^2 + 6x - 16} \over {x^2 + x -6}}, x \neq 2} \right\}\] \[{7x - 4}, x =2\]

  2. anonymous
    • 5 years ago
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    I can't order the brackets

  3. myininaya
    • 5 years ago
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    x^2+x-6=(x+3)(x-2) x^2+6x-16 doesn't have either factor so the limit does not exsit at -3 and 2 so the function is not continuous at either x

  4. myininaya
    • 5 years ago
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    oh x=2 is defined for 7x-4

  5. myininaya
    • 5 years ago
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    still since the limit doesn't exist at x=2 then f is not continuous there

  6. anonymous
    • 5 years ago
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    yeap. sorry for that.

  7. anonymous
    • 5 years ago
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    Oh! OK! Thanks.

  8. myininaya
    • 5 years ago
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    all you have to do is make sure the limit exists and the limx->af(x)=f(a) then f is continuous at x=a

  9. nikvist
    • 5 years ago
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    \[\lim_{x\rightarrow 2}f(x)=\lim_{x\rightarrow 2}\frac{x^2+6x-16}{x^2+x-6}=\lim_{x\rightarrow 2}\frac{(x+8)(x-2)}{(x+3)(x-2)}=\lim_{x\rightarrow 2}\frac{x+8}{x+3}=2\] \[f(2)=7\cdot 2-4=10\] \[\lim_{x\rightarrow 2}f(x)\neq f(2)\Rightarrow f(x)\quad not\enspace continuous\]

  10. anonymous
    • 5 years ago
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    Thanks!

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