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Continuity And Differentiability

Question

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Prove that the function f(x) = 5 x – 3 is continuous at x = 0, at x = – 3 and at x = 5.

Solution

Here space straight f left parenthesis straight x right parenthesis equals 5 straight x minus 3 left parenthesis straight i right parenthesis space Lt with straight x rightwards arrow 0 below space straight f left parenthesis straight x right parenthesis equals stack space Lt with straight x rightwards arrow 0 below space left parenthesis 5 straight x minus 3 right parenthesis equals 5 left parenthesis 0 right parenthesis minus 3 equals 0 minus 3 equals negative 3 Now space space straight f space is space defined space at space straight x equals 0 and space space straight f left parenthesis 0 right parenthesis equals 5 left parenthesis 0 right parenthesis minus 3 equals 0 minus 3 equals negative 3 therefore space Lt with straight x rightwards arrow 0 below space straight f left parenthesis straight x right parenthesis equals space straight f left parenthesis 0 right parenthesis equals negative 3 therefore space straight f space is space continous space at space straight x equals 0.

left parenthesis ii right parenthesis space stack Lt space with straight x rightwards arrow negative 3 below space straight f left parenthesis straight x right parenthesis equals stack space Lt with straight x rightwards arrow negative 3 below space left parenthesis 5 straight x minus 3 right parenthesis equals 5 left parenthesis negative 3 right parenthesis minus 3 equals negative 15 minus 3 equals negative 18 Now space straight f space is space defined space at space straight x equals negative 3 and space space straight f left parenthesis negative 3 right parenthesis equals 5 left parenthesis negative 3 right parenthesis minus 3 equals negative 15 minus 3 equals negative 18 therefore stack space Lt space with straight x rightwards arrow negative 3 below space straight f left parenthesis straight x right parenthesis equals straight f left parenthesis negative 3 right parenthesis equals negative 18 therefore space straight f space is space continous space at space straight x equals negative 3.

left parenthesis iii right parenthesis stack space Lt space with straight x rightwards arrow 5 below straight f left parenthesis straight x right parenthesis equals stack space Lt with straight x rightwards arrow 5 below space left parenthesis 5 straight x minus 3 right parenthesis equals 5 left parenthesis 5 right parenthesis minus 3 equals 25 minus 3 equals 22 Now space straight f space is space defined space at space straight x equals 5 and space space straight f left parenthesis 5 right parenthesis equals 5 left parenthesis 5 right parenthesis minus 3 equals 25 minus 3 equals 22 therefore Lt with straight x rightwards arrow 5 below space straight f left parenthesis straight x right parenthesis equals straight f left parenthesis 5 right parenthesis equals 22 therefore straight f space is space continous space at space straight x equals 5.

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