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mathcalculus Group Title

HELP: CONTINUITY B=? (attached below)

  • one year ago
  • one year ago

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  1. mathcalculus Group Title
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    • one year ago
  2. pgpilot326 Group Title
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    plug in x = 3. You'll get a number for the first function and set the second function equal to that value. Plug in 3 for x and solve for b.

    • one year ago
  3. pgpilot326 Group Title
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    in the second function

    • one year ago
  4. mathcalculus Group Title
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    i did i got it wrong.. :/

    • one year ago
  5. mathcalculus Group Title
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    can you show me how you would do it?

    • one year ago
  6. mathcalculus Group Title
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    @pgpilot326

    • one year ago
  7. pgpilot326 Group Title
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    From the first equation:\[4(3)-6=12-6=6\] Thus -7(x)+b = 6 when x=3. So \[-7(3)+b=6 \rightarrow -12+b=6 \rightarrow b=21+6=27\] Is this what you got?

    • one year ago
  8. mathcalculus Group Title
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    yes, i missed one step, thank you.

    • one year ago
  9. mathcalculus Group Title
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    I got it, thanks alot.

    • one year ago
  10. surjithayer Group Title
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    Please note a function is continuos at x=a, if (1) L.H.L=R.H.L (2) \[\lim_{x \rightarrow a} f(x)=f(a)\] \[\lim_{x \rightarrow 3-}f \left( x \right)=\lim_{x \rightarrow3-}\left( 4x-6 \right)\] \[put x=3-h,h>0,h \rightarrow 0 as x \rightarrow 3-\] \[\lim_{x \rightarrow 3-}f \left( x \right)=\lim_{h \rightarrow 0}\left\{ 4\left( 3-h \right)-6 \right\}=4\left( 3-0 \right)-6=6 \] \[\lim_{x \rightarrow 3+}f \left( x \right)=\lim_{x \rightarrow 3+} \left( -7x+b \right)\] \[put x=3+h,h \rightarrow 0,as x \rightarrow 3+\] \[\lim_{x \rightarrow 3+} f \left( x \right)=\lim_{h \rightarrow 0} \left\{- 7\left( 3+h \right)+b \right\}\]=-7(3+0)+b=-21+b Because f(x) is continuous at all real values \[Hence \lim_{x \rightarrow3-}f(x)=\lim_{x \rightarrow 3+} f(x)\] 6=-21+b b=6+21=27

    • one year ago
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