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Application Of Derivatives

Question
CBSEENMA12035295

Let I be any interval disjoint from (– 1, 1). Prove that the function f given by straight f apostrophe left parenthesis straight x right parenthesis space equals space straight x plus 1 over straight x is strictly increasing on I.

Solution

Here straight f left parenthesis straight x right parenthesis space equals space straight x plus 1 over straight x space space space space space space rightwards double arrow space space space space straight f apostrophe left parenthesis straight x right parenthesis space equals space 1 minus 1 over straight x squared space equals space fraction numerator straight x squared minus 1 over denominator straight x squared end fraction
Now, straight f apostrophe left parenthesis straight x right parenthesis greater than 0 space space if space fraction numerator straight x squared minus 1 over denominator straight x squared end fraction greater than 0
space straight i. straight e. comma space space space if space straight x squared minus 1 greater than 0 space space space space straight i. straight e. comma space space if space straight x squared greater than 1 space space space space straight i. straight e. comma space if space open vertical bar straight x squared close vertical bar greater than 1
straight i. straight e. comma space space space space if space open vertical bar straight x close vertical bar greater than 1 space space space space straight i. straight e. space if space either space straight x less than negative 1 space space space or space space straight x greater than 1
straight i. straight e. comma space space if space straight x element of space left parenthesis negative infinity comma space minus 1 right parenthesis space space or space space space straight x space space element of space left parenthesis 1 comma space infinity right parenthesis
straight i. straight e. comma space if space straight x space element of space left parenthesis negative infinity comma space minus 1 right parenthesis space space space union space space space space left parenthesis 1 comma space infinity right parenthesis
therefore space space space straight f left parenthesis straight x right parenthesis space is space strictly space increasing space on space 1.