Question

Which of the following functions are continuous at the indicated points ?
$straight f left parenthesis straight x right parenthesis equals open curly brackets table attributes columnalign left end attributes row cell 2 straight x minus 1 space space space comma space space space if space straight x less than 0 end cell row cell 2 straight x plus 6 space space space comma space space space if space straight x greater than 1 end cell end table close at space straight x equals 0$

Solution

straight f left parenthesis straight x right parenthesis equals open curly brackets table attributes columnalign left end attributes row cell 2 straight x minus 1 space space space comma space space space if space straight x less than 0 end cell row cell 2 straight x plus 6 space space space comma space space space if space straight x greater than 1 end cell end table close space Lt with straight x rightwards arrow 0 to the power of minus below straight f left parenthesis straight x right parenthesis equals Lt with straight x rightwards arrow 0 to the power of minus below left parenthesis 2 straight x minus 1 right parenthesis equals Lt with straight h rightwards arrow 0 below left curly bracket 2 left parenthesis 0 minus straight h right parenthesis minus 1 right curly bracket space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space left square bracket Put space straight x equals 0 minus straight h comma space straight h greater than 0 space so space that space straight h rightwards arrow 0 space as space straight x rightwards arrow 0 right square bracket space space space space space space space space space space space space space space space space space equals Lt left parenthesis negative 2 straight h minus 1 right parenthesis equals 0 minus 1 equals negative 1

But f (0) = Value of (2 x + 6) at .x = 0
= 2 x 0 + 6 = 0 + 6 = 6

therefore space Lt with straight x rightwards arrow 0 to the power of minus below straight f left parenthesis straight x right parenthesis not equal to straight f left parenthesis 0 right parenthesis

∴ f(x) is discontinuous at x = 0.

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