Question

Please see attachment

Answer 1

Answer:

a) The value of absolute minimum value = - 0.3536  

b) which is attained at   $x = \frac{1}{\sqrt{2} }$  

Step-by-step explanation:

Step(i):-

Given function

                       $f(x) = \frac{-x}{2x^{2} +1}$     ...(i)

Differentiating equation (i) with respective to 'x'

                     $f^{l} = \frac{2x^{2} +1(-1) - (-x) (4x)}{(2x^{2}+1)^{2} }$   ...(ii)

                    $f^{l}(x) = \frac{2x^{2}-1}{(2x^{2}+1)^{2} }$

Equating Zero

                   $f^{l}(x) = \frac{2x^{2}-1}{(2x^{2}+1)^{2} } = 0$

                 $\frac{2x^{2}-1}{(2x^{2}+1)^{2} } = 0$

                $2 x^{2}-1 = 0$

               $2 x^{2} = 1$

             $x^{2} = \frac{1}{2}$

             $x = \frac{-1}{\sqrt{2} } , x = \frac{1}{\sqrt{2} }$

Step(ii):-

Again Differentiating equation (ii) with respective to 'x'

$f^{ll}(x) = \frac{(2x^{2} +1)^{2} (4x) - 2(2x^{2} +1) (4x)(2x^{2}-1) }{(2x^{2}+1)^{4} }$

put

      $x = \frac{1}{\sqrt{2} }$

$f^{ll} (x) > 0$

The absolute minimum value at   $x = \frac{1}{\sqrt{2} }$

Step(iii):-

The value of absolute minimum value

                         $f(x) = \frac{-x}{2x^{2} +1}$

                       $f(\frac{1}{\sqrt{2} } ) = \frac{-\frac{1}{\sqrt{2} } }{2(\frac{1}{\sqrt{2} } )^{2} +1}$

         on calculation we get

The value of absolute minimum value = - 0.3536      

Final answer:-

a) The value of absolute minimum value = - 0.3536  

b) which is attained at   $x = \frac{1}{\sqrt{2} }$