Maths-
General
Easy

Question

The function x square root of 1 minus x to the power of 2 end exponent end root comma left parenthesis x greater than 0 right parenthesis has

  1. A local maxima  
  2. A local minima  
  3. Neither a local maxima nor a local minima  
  4. None of the above  

The correct answer is: A local maxima


    Let f open parentheses x close parentheses equals x square root of 1 minus x to the power of 2 end exponent end root
    rightwards double arrow blank f to the power of ´ end exponent open parentheses x close parentheses equals fraction numerator 1 minus 2 x to the power of 2 end exponent over denominator square root of 1 minus x to the power of 2 end exponent end root end fraction

    rightwards double arrow blank For maxima or minima, put fraction numerator d y over denominator d x end fraction equals 0

    x equals plus-or-minus fraction numerator 1 over denominator square root of 2 end fraction

    But x greater than 0, therefore we have x equals fraction numerator 1 over denominator square root of 2 blank end root end fraction

    Now, f to the power of ´ ´ end exponent open parentheses x close parentheses equals fraction numerator square root of 1 minus x to the power of 2 end exponent end root open parentheses negative 4 x close parentheses minus open parentheses 1 minus 2 x to the power of 2 end exponent close parentheses fraction numerator negative x over denominator square root of 1 minus x to the power of 2 end exponent end root end fraction over denominator left parenthesis 1 minus x to the power of 2 end exponent right parenthesis end fraction

    equals fraction numerator 2 x to the power of 3 end exponent minus 3 x over denominator open parentheses 1 minus x to the power of 2 end exponent close parentheses to the power of 3 divided by 2 end exponent end fraction

    rightwards double arrow blank f ´ ´ open parentheses fraction numerator 1 over denominator square root of 2 end fraction close parentheses=minusve, maximum.

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