Discuss continuity of the function f given by

f(x) = | x – 1| + | x – 2 ] at x = 1 and x = 2.

Here space straight f left parenthesis straight x equals open vertical bar straight x minus 1 close vertical bar plus open vertical bar straight x minus 2 close vertical bar right parenthesis 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 open parentheses open vertical bar straight x minus 1 close vertical bar plus open vertical bar straight x minus 2 close vertical bar close parentheses 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 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 0 below open parentheses open vertical bar 1 minus straight h plus 1 close vertical bar plus open vertical bar 1 minus straight h minus 2 close vertical bar close parentheses 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 h rightwards arrow 0 below left parenthesis open vertical bar negative straight h close vertical bar plus open vertical bar negative 1 minus straight h close vertical bar right parenthesis space Lt with straight x rightwards arrow 1 to the power of minus below straight f left parenthesis straight x right parenthesis equals left parenthesis straight h plus 1 plus straight h right parenthesis equals 0 plus 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 minus 1 close vertical bar plus open vertical bar straight x minus 2 close vertical bar right parenthesis 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 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 minus 1 close vertical bar plus open vertical bar 1 plus straight h minus 2 close vertical bar equals Lt with straight x rightwards arrow 0 below left parenthesis open vertical bar straight h close vertical bar plus open vertical bar left parenthesis 1 minus straight h right parenthesis 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 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 1 right parenthesis equals open vertical bar 1 minus 1 close vertical bar plus open vertical bar 1 minus 2 close vertical bar equals 0 plus 1 equals 1 therefore 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 plus below straight f left parenthesis straight x right parenthesis equals straight f left parenthesis 1 right parenthesis therefore straight f space is space continous space at space straight x equals 1.

∴ f is continuous at .x = 1

space Lt with straight x rightwards arrow 2 to the power of minus below straight f left parenthesis straight x right parenthesis equals Lt with straight x rightwards arrow 2 to the power of minus below left parenthesis open vertical bar straight x minus 1 close vertical bar plus open vertical bar straight x plus 2 close vertical bar right parenthesis space space space space space space space space space space space space space space space space left square bracket Put space straight x equals 2 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 equals Lt with straight h rightwards arrow 0 below left parenthesis open vertical bar 2 minus straight h minus 1 close vertical bar plus open vertical bar 2 minus straight h minus 2 close vertical bar right parenthesis 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 equals Lt with straight h rightwards arrow 0 below left parenthesis 1 minus straight h plus straight h right parenthesis equals 1 minus 0 plus 0 equals 1 space Lt with straight x rightwards arrow 2 to the power of plus below straight f left parenthesis straight x right parenthesis equals Lt with straight x rightwards arrow 2 to the power of plus below straight f left parenthesis open vertical bar straight x minus 1 close vertical bar plus open vertical bar straight x minus 2 close vertical bar right parenthesis space space space space space space space space space space space space space space space left square bracket Put space straight x equals 2 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 space equals Lt with straight h rightwards arrow 0 below open curly brackets open vertical bar 2 plus straight h minus 1 close vertical bar plus open vertical bar 2 plus straight h minus 2 close vertical bar close curly brackets 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 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

Also space straight f left parenthesis 2 right parenthesis equals open vertical bar 2 minus 1 close vertical bar plus open vertical bar 2 minus 2 close vertical bar equals 1 plus 0 equals 1 therefore Lt with straight x rightwards arrow 2 to the power of minus below straight f left parenthesis straight x right parenthesis equals Lt with straight x rightwards arrow 2 to the power of plus below straight f left parenthesis straight x right parenthesis equals straight f left parenthesis 2 right parenthesis

∴ f is continuous at x = 2.

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Discuss continuity of the function f given by f(x) = | x – 1| + | x – 2 ] at x = 1 and x = 2.

Continuity and Differentiability

Mathematics Part I

Discuss continuity of the function f given by

f(x) = | x – 1| + | x – 2 ] at x = 1 and x = 2.