Question

Find the value of a and b so that the function f given by
$straight f left parenthesis straight x right parenthesis equals open curly brackets table row cell 1 space space space comma space space space space space straight x less or equal than 3 end cell row cell ax plus straight b comma space space space space 3 less than straight x less than 5 end cell row cell 7 space space space comma space space space space straight x greater or equal than 5 end cell end table close$
is continous at x=3 and x=5

Solution

Here space straight f left parenthesis straight x right parenthesis equals open curly brackets table row cell 1 space space space comma space space space space space straight x less or equal than 3 end cell row cell ax plus straight b comma space space space space 3 less than straight x less than 5 end cell row cell 7 space space space comma space space space space straight x greater or equal than 5 end cell end table close

∵ f is continuous at x = 3 and x = 5

∴ f is right continuous at x = 3 and left continuous at x = 5

therefore space Lt with straight x rightwards arrow 3 to the power of plus below straight f left parenthesis straight x right parenthesis equals straight f left parenthesis 3 right parenthesis space space space space space space space space space space space rightwards double arrow space 3 straight a plus straight b equals 1 space space space space space space space.... left parenthesis 1 right parenthesis space Lt with straight x rightwards arrow 5 to the power of minus below space space space space straight f left parenthesis straight x right parenthesis equals straight f left parenthesis 5 right parenthesis space space space space space space space space space space space space rightwards double arrow space 5 straight a plus straight b equals 7 space space space space space space space.... left parenthesis 2 right parenthesis

Subtracting (2) from (1), we get,
– 2 a = – 6, ⇒ a = 3
Putting a = 3 in (1) we get,
3 x 3 + b = 1, ⇒ b = – 8
∴ a = 3, b = –8.

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