Limits and Derivatives

Let find a so that f may be derivable at x =1. - Wired Faculty

Question

Let $space space space straight f left parenthesis straight x right parenthesis space equals space open curly brackets table attributes columnalign left end attributes row cell straight x plus straight a comma space straight x less or equal than 1 end cell row cell ax squared plus 1 comma space straight x greater than 1 end cell end table close$

find a so that f may be derivable at x =1.

Solution

Here, f(1) = 1+a
$space space space space space space space space space space space Lf apostrophe left parenthesis 1 right parenthesis space equals space limit as straight h rightwards arrow 0 to the power of minus of fraction numerator straight f left parenthesis 1 plus straight h right parenthesis minus straight f left parenthesis 1 right parenthesis over denominator straight h end fraction equals limit as straight h rightwards arrow 0 to the power of minus of fraction numerator left parenthesis 1 plus straight h right parenthesis plus straight a minus left parenthesis 1 plus straight a right parenthesis over denominator straight h end fraction$
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$space space space space space space Rf apostrophe left parenthesis 1 right parenthesis space space equals space limit as straight h rightwards arrow 0 to the power of plus of fraction numerator straight f left parenthesis 1 plus straight h right parenthesis minus straight f left parenthesis 1 right parenthesis over denominator straight h end fraction$ (here h >0 $rightwards double arrow$ 1 + h > 1)

$space space space space space space equals space limit as straight h rightwards arrow 0 to the power of plus of fraction numerator straight a left parenthesis 1 plus straight h right parenthesis squared plus 1 minus left parenthesis 1 plus straight a right parenthesis over denominator straight h end fraction$

 $space space space space space space space space space equals space limit as straight h rightwards arrow 0 to the power of plus of fraction numerator ah squared plus 2 ah over denominator straight h end fraction equals limit as straight h rightwards arrow 0 to the power of plus of left parenthesis ah plus 2 straight a right parenthesis equals 0 plus 2 straight a equals 2 straight a$
Since f is derivable at x =1

∴ Lf'(1) = Rf'(1)
$rightwards double arrow$ 1 = 2a $rightwards double arrow$    $space space space space straight a space equals space 1 half$

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