Let [f( x ) = x^2 + 1 x^2] and [g( x ) = x - 1 x], [x R - - 1,0,1 ].

Question

Let $$f\left( x \right) = {x^2} + {1 \over {{x^2}}}$$ and $$g\left( x \right) = x - {1 \over x}$$,

$$x \in R - \left\{ { - 1,0,1} \right\}$$.

If $$h\left( x \right) = {{f\left( x \right)} \over {g\left( x \right)}}$$, then the local minimum value of h(x) is

Accepted Answer

Given $$f\left( x \right) = {x^2} + {1 \over {{x^2}}}$$ and $$g\left( x \right) = x - {1 \over x}$$

As $$h\left( x \right) = {{f\left( x \right)} \over {g\left( x \right)}}$$

= $${{{x^2} + {1 \over {{x^2}}}} \over {x - {1 \over x}}}$$

= $${{{{\left( {x - {1 \over x}} \right)}^2} + 2.x.{1 \over x}} \over {x - {1 \over x}}}$$

= $${{{{\left( {x - {1 \over x}} \right)}^2} + 2} \over {x - {1 \over x}}}$$

Let $${x - {1 \over x} = t}$$

So $$f\left( t \right) = {{{t^2} + 2} \over t}$$ = $$t + {2 \over t}$$

then $$f'\left( t \right) = 1 - {2 \over {{t^2}}}$$

At maximum or minimum $$f'\left( t \right) = 0$$.

$$\therefore$$ $$1 - {2 \over {{t^2}}} = 0$$

$$ \Rightarrow t =  \pm \sqrt 2 $$

Now we need to find $$f''\left( t \right)$$ which will tell at which point among $$ + \sqrt 2 $$ and $$ - \sqrt 2 $$ will be maximum and minimum.

$$f''\left( t \right) =  + {4 \over {{t^3}}}$$

So when $$t =  + \sqrt 2 $$, then $$f''\left( t \right) =  + {4 \over {{{\left( {\sqrt 2 } \right)}^3}}}$$ = $$ + \sqrt 2 $$

As $$f''\left( t \right) > 0$$ when $$t =  + \sqrt 2 $$ then at $$t =  + \sqrt 2 $$ the function $$f\left( t \right)$$ will be minimum.

Min of $$f\left( t \right) = {{{{\left( {\sqrt 2 } \right)}^2} + 2} \over {\sqrt 2 }}$$ = $${4 \over {\sqrt 2 }}$$ = $$2\sqrt 2 $$

So local minimum of $$f\left( t \right)$$ = $$2\sqrt 2 $$

Solutions · 1