anonymous
  • anonymous
For what value of the constant c is the function f continuous on ( -\infty , \infty ) where f( t ) = \begin{cases} { t }^2 - c & \text{ if } t \in (-\infty, 2) \quad \quad \\ c { t } + 8 & \text{ if }t \in [ 2 , \infty) \quad \quad \end{cases} c= ?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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eyust707
  • eyust707
in order to make them continous they both need to be equal when the functions change. so basically they need to be equal when t is 2
anonymous
  • anonymous
ok
anonymous
  • anonymous
and then i solve the equations?

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apoorvk
  • apoorvk
okay so for the function to be continuous in all R, at t=2, the two different part of the function will have to be equal. so, t^2-c=ct+8 put t=2, simplify and get value of c. then enjoy!!
eyust707
  • eyust707
\[4- c = 2c + 8\]
anonymous
  • anonymous
-4/3
anonymous
  • anonymous
thanks
eyust707
  • eyust707
correct!

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