Question 1187848
 {{{f(x)= x/8 + 3/(2x) }}}

A turning point is a point of the graph where the graph changes from increasing to decreasing (rising to falling) or decreasing to increasing (falling to rising). 

(i)

to find them, derivate

{{{f}}}'{{{(x) = 1/8 - 3/(2x^2)}}}

equal to zero and solve for{{{ x}}}

{{{0= 1/8 - 3/(2x^2)}}}

{{{3/(2x^2)= 1/8}}}.........cross multiply

{{{24= 2x^2}}}

{{{x^2=12}}}

{{{x=sqrt(12))}}}

{{{x=sqrt(4*3)}}}

{{{x}}}=±{{{2sqrt(3)}}}

solutions:

{{{x=2sqrt(3)}}}  or {{{x=-2sqrt(3)}}}


substitute in {{{f(x)= x/8 + 3/(2x) }}}


{{{f(x)= (2sqrt(3))/8 + 3/(2*2sqrt(3)) }}}

{{{f(x)= sqrt(3)/2}}}

or

{{{f(x)= (-2sqrt(3))/8 + 3/(-2*2sqrt(3))}}} 

{{{f(x)= -sqrt(3)/2}}}


turning points are at:

({{{2sqrt(3)}}}, {{{sqrt(3)/2}}}) 
({{{-2sqrt(3)}}}, {{{-sqrt(3)/2}}}) 

(ii) Determine  the nature  of the stationary points

({{{2sqrt(3)}}}, {{{sqrt(3)/2}}})->Minimum
({{{-2sqrt(3)}}}, {{{-sqrt(3)/2}}})->Maximum

(iii)

{{{ drawing( 600, 600, -10, 10, -10, 10,
circle(2sqrt(3), sqrt(3)/2,.12), circle(-2sqrt(3), -sqrt(3)/2,.12), 
locate(2sqrt(3), sqrt(3)/2,p(2sqrt(3), sqrt(3)/2)), 
locate(-2sqrt(3), -sqrt(3)/2,p(-2sqrt(3), -sqrt(3)/2)), 
graph( 600, 600, -10, 10, -10, 10, x/8 + 3/(2x) )) }}}