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Continuity And Differentiability

Question

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Show that the function f given by f(x) = | x | + | x –1 |,x ∈ R is continuous both at x =.0 and x = 1.

Solution

Here space straight f left parenthesis straight x right parenthesis equals open vertical bar straight x close vertical bar plus open vertical bar straight x minus 1 close vertical bar 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 open vertical bar straight x close vertical bar plus open vertical bar straight x minus 1 close vertical bar right parenthesis 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 right square bracket space space space space space space space space space space space space space space space space equals Lt with straight h rightwards arrow 0 below left parenthesis open vertical bar 0 minus straight h close vertical bar plus open vertical bar 0 minus straight h minus 1 close vertical bar right parenthesis equals Lt with straight h rightwards arrow 0 below open curly brackets straight h plus left parenthesis 1 plus straight h right parenthesis close curly brackets equals 0 plus 1 plus 0 equals 1 space Lt with straight x rightwards arrow 0 to the power of plus below straight f left parenthesis straight x right parenthesis equals Lt with straight x rightwards arrow 0 to the power of plus below left parenthesis open vertical bar straight x close vertical bar plus open vertical bar straight x minus 1 close vertical bar right parenthesis 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 plus straight h comma space straight h greater than 0 right square bracket space space space space space space space space space space space space space space space space equals Lt with straight h rightwards arrow 0 below left parenthesis open vertical bar 0 plus straight h close vertical bar plus open vertical bar 0 plus straight h minus 1 close vertical bar right parenthesis equals Lt with straight h rightwards arrow 0 below open curly brackets open vertical bar straight h close vertical bar plus open vertical bar negative left parenthesis 1 minus straight h right parenthesis close vertical bar close curly brackets space space space space space space space space space space space space space space space space equals Lt with straight h rightwards arrow 0 below left parenthesis straight h plus 1 minus straight h right parenthesis equals Lt with straight h rightwards arrow 0 below left parenthesis straight h plus 1 minus straight h right parenthesis equals Lt with straight h rightwards arrow 0 below 1 equals 1 Also space straight f left parenthesis 0 right parenthesis space equals open vertical bar 0 close vertical bar plus open vertical bar 0 plus 1 close vertical bar equals 0 plus 1 equals 1 therefore 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 plus below straight f left parenthesis straight x right parenthesis equals straight F left parenthesis 0 right parenthesis therefore straight f space is space continous space at space straight x equals 0. Again space Lt with straight x rightwards arrow 1 to the power of minus below straight f left parenthesis straight x right parenthesis equals Lt with straight x rightwards arrow 1 to the power of minus below left parenthesis open vertical bar straight x close vertical bar plus open vertical bar straight x minus 1 close vertical bar right parenthesis 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 1 minus straight h comma space straight h greater than 0 right square 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 equals Lt with straight h rightwards arrow 0 below open vertical bar 1 minus straight h close vertical bar plus open vertical bar 1 minus straight h minus 1 close vertical bar equals Lt with straight h rightwards arrow 0 below left parenthesis open vertical bar 1 minus straight h close vertical bar plus open vertical bar negative straight h close vertical bar right parenthesis 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 equals Lt with straight h rightwards arrow 0 below left parenthesis 1 minus straight h plus straight h right parenthesis equals Lt with straight h rightwards arrow 0 below 1 equals 1 space Lt with straight x rightwards arrow 1 to the power of plus below straight f left parenthesis straight x right parenthesis equals Lt with straight x rightwards arrow 1 to the power of plus below left parenthesis open vertical bar straight x close vertical bar plus open vertical bar straight x minus 1 close vertical bar right parenthesis space space space left square bracket Put space straight x equals 1 plus straight h comma space straight h greater than 0 right square bracket space space space space space space space space space space space space space space space space equals Lt with straight h rightwards arrow 0 below left parenthesis open vertical bar 1 plus straight h close vertical bar plus open vertical bar 1 plus straight h minus 1 close vertical bar right parenthesis equals Lt with straight h rightwards arrow 0 below left parenthesis open vertical bar 1 plus straight h close vertical bar plus open vertical bar straight h close vertical bar right parenthesis space space space space space space space space space space space space space space space space equals Lt with straight h rightwards arrow 0 below left parenthesis 1 plus straight h plus straight h right parenthesis equals 1 plus 0 plus 0 equals 1 straight f left parenthesis 1 right parenthesis equals open vertical bar 1 close vertical bar plus open vertical bar 1 minus 1 close vertical bar equals 1 plus 0 equals 1 therefore Lt with straight x rightwards arrow 1 to the power of minus below straight f left parenthesis straight x right parenthesis equals Lt with straight x rightwards arrow 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.

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