Question 1178955


 {{{f(x)=16x/(x^2+2)}}}

 {{{h(x)=12x/(x^2+1)}}}


1 when are equal

 {{{16x/(x^2+2)=12x/(x^2+1)}}}

{{{ 16x(x^2+1)=12x(x^2+2)}}}..........divide by {{{x}}}  (remmember count {{{x=0}}} as a solution)

{{{16(x^2+1)=12(x^2+2)}}}

{{{16x^2+16=12x^2+24}}}

{{{16x^2+16-12x^2-24=0}}}

{{{4x^2-8=0}}}

{{{x^2-2=0}}}

{{{x^2=2}}}


solutions:

{{{x=-sqrt(2)}}}

{{{x=sqrt(2)}}}

{{{x=0 }}}

2.

{{{f}}}'{{{(x) = -(16 (x^2 - 2))/(x^2 + 2)^2}}}

{{{h}}}'{{{(x) = -(12 (x^2 - 1))/(x^2 + 1)^2}}}


when greater than the actual function:


{{{16x/(x^2+2)  <-(16 (x^2 - 2))/(x^2 + 2)^2 }}}

for all {{{x < 0.650629}}},  in interval

({{{-infinity}}},{{{ 0.650629}}})


{{{12x/(x^2+1) < -(12 (x^2 - 1))/(x^2 + 1)^2}}}

for all {{{x < 0.543689}}},  in interval

({{{-infinity}}}, {{{0.543689}}})