Answer:
a) The value of absolute minimum value = - 0.3536
b) which is attained at [tex]x = \frac{1}{\sqrt{2} }[/tex]
Step-by-step explanation:
Step(i):-
Given function
[tex]f(x) = \frac{-x}{2x^{2} +1}[/tex] ...(i)
Differentiating equation (i) with respective to 'x'
[tex]f^{l} = \frac{2x^{2} +1(-1) - (-x) (4x)}{(2x^{2}+1)^{2} }[/tex] ...(ii)
[tex]f^{l}(x) = \frac{2x^{2}-1}{(2x^{2}+1)^{2} }[/tex]
Equating Zero
[tex]f^{l}(x) = \frac{2x^{2}-1}{(2x^{2}+1)^{2} } = 0[/tex]
[tex]\frac{2x^{2}-1}{(2x^{2}+1)^{2} } = 0[/tex]
[tex]2 x^{2}-1 = 0[/tex]
[tex]2 x^{2} = 1[/tex]
[tex]x^{2} = \frac{1}{2}[/tex]
[tex]x = \frac{-1}{\sqrt{2} } , x = \frac{1}{\sqrt{2} }[/tex]
Step(ii):-
Again Differentiating equation (ii) with respective to 'x'
[tex]f^{ll}(x) = \frac{(2x^{2} +1)^{2} (4x) - 2(2x^{2} +1) (4x)(2x^{2}-1) }{(2x^{2}+1)^{4} }[/tex]
put
[tex]x = \frac{1}{\sqrt{2} }[/tex]
[tex]f^{ll} (x) > 0[/tex]
The absolute minimum value at [tex]x = \frac{1}{\sqrt{2} }[/tex]
Step(iii):-
The value of absolute minimum value
[tex]f(x) = \frac{-x}{2x^{2} +1}[/tex]
[tex]f(\frac{1}{\sqrt{2} } ) = \frac{-\frac{1}{\sqrt{2} } }{2(\frac{1}{\sqrt{2} } )^{2} +1}[/tex]
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 [tex]x = \frac{1}{\sqrt{2} }[/tex]
