Question

Discuss the continuity of the function f , where fis defined by
$straight f left parenthesis straight x right parenthesis equals open curly brackets table row cell negative 2 comma space space space space if space straight x less or equal than negative 1 end cell row cell 2 straight x comma space space if minus 1 less than straight x less or equal than 1 end cell row cell 2 space space space space space if space straight x greater than 1 end cell end table close$

Solution

$straight f left parenthesis straight x right parenthesis equals open curly brackets table row cell negative 2 comma space space space space if space straight x less or equal than negative 1 end cell row cell 2 straight x comma space space if minus 1 less than straight x less or equal than 1 end cell row cell 2 space space space space space if space straight x greater than 1 end cell end table close$
Function f is defined at all points of the real line.
When x < – 1, we have f(x) = – 2; which is constant and so it is continuous.
$At space straight x equals negative 1 space space space space space space space space Lt with straight x rightwards arrow negative 1 to the power of minus below straight f left parenthesis straight x right parenthesis equals Lt with straight x rightwards arrow negative 1 to the power of minus below left parenthesis negative 2 right parenthesis equals negative 2 space space space space space space space space Lt with straight x rightwards arrow negative 1 to the power of plus below straight f left parenthesis straight x right parenthesis equals Lt with straight x rightwards arrow negative 1 to the power of plus below left parenthesis 2 straight x right parenthesis equals 2 left parenthesis negative 1 right parenthesis equals negative 2 Also space space space space space space space space straight f left parenthesis negative 1 right parenthesis equals negative 2 therefore space stack space space space space space space space space space space Lt with straight x rightwards arrow negative 1 to the power of minus below straight f left parenthesis straight x right parenthesis equals Lt with straight x rightwards arrow negative 1 to the power of plus below straight f left parenthesis straight x right parenthesis equals straight f left parenthesis negative 1 right parenthesis$

∴ f is continuous at x = – 1
In the interval –1 < x < 1, we have f(x) = 2 x, which being a linear polynomial, is continuous.
$At space straight x equals negative 1 space space space space space space space space Lt with straight x rightwards arrow negative 1 to the power of minus below straight f left parenthesis straight x right parenthesis equals Lt with straight x rightwards arrow negative 1 to the power of minus below left parenthesis 2 straight x right parenthesis equals 2 left parenthesis 1 right parenthesis equals 2 space space space space space space space space Lt with straight x rightwards arrow negative 1 to the power of plus below straight f left parenthesis straight x right parenthesis equals Lt with straight x rightwards arrow negative 1 to the power of plus below left parenthesis 2 right parenthesis equals 2 Also space space space space space space space space straight f left parenthesis 1 right parenthesis equals 2 left parenthesis 1 right parenthesis equals 2 therefore space stack space space space space space space space space space space Lt space with straight x rightwards arrow negative 1 to the power of minus below straight f left parenthesis straight x right parenthesis equals Lt with straight x rightwards arrow negative 1 to the power of plus below straight f left parenthesis straight x right parenthesis equals straight f left parenthesis 1 right parenthesis$
∴ f is continuous at x = 1
When x > 1, we have f (x) = 2. which is constant and so it is continuous.

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