Question

Find the value of k so that the function f is continuous at the indicated point
$straight f left parenthesis straight x right parenthesis equals open curly brackets table attributes columnalign left end attributes row cell straight k space straight x plus 1 space space comma space straight x less or equal than 5 end cell row cell 3 x minus 5 space space space comma space straight x greater than 5 end cell end table close at space straight x equals 5$

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Solution

Here space straight f left parenthesis straight x right parenthesis equals open curly brackets table attributes columnalign left end attributes row cell straight k space straight x plus 1 space space comma space straight x less or equal than 5 end cell row cell 3 straight x minus 5 space space space comma space straight x greater than 5 end cell end table close space Lt with straight x rightwards arrow 5 to the power of minus below straight f left parenthesis straight x right parenthesis equals Lt with straight x rightwards arrow 5 to the power of minus below left parenthesis straight k space straight x plus 1 right parenthesis equals 5 straight k plus 1 space Lt with straight x rightwards arrow 5 to the power of plus below straight f left parenthesis straight x right parenthesis equals Lt with straight x rightwards arrow 5 to the power of plus below left parenthesis 3 straight x minus 5 right parenthesis equals 15 minus 5 equals 10 Since space straight f space is space continous space at space straight x equals 1 therefore space Lt with straight x rightwards arrow 5 to the power of minus below straight f left parenthesis straight x right parenthesis equals Lt with straight x rightwards arrow 5 to the power of plus below straight f left parenthesis straight x right parenthesis therefore space space 5 straight k plus 1 equals 10 space rightwards double arrow space 5 straight k equals 9 space rightwards double arrow space straight k equals 9 over 5

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