How to find p ?

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Answer 1 · 2018-05-25T13:35:39

Please see below.

If the gradient has a turning point at #(p,3)#, then #(8-x^3)/(2x^5)# must equal to #0#.

Sub #x=p#

#(8-p^3)/(2p^5)=0#

#8-p^3=0#

#8=p^3#

#p=2#

(i) Answer for #p# is #2#

#int (8-x^3)/(2x^5) dx#

= #int(4/x^5-1/(2x^2)) dx#

= #4times(-1/4)times1/x^4-1/2times(-1)times1/x+c#

= #-1/x^4+1/(2x)+c#

Sub in #(2,3)#

#y=-1/2^4+1/(2*2)+c#

#3=-1/16+1/4+c#

#c=3-3/16#

#c=45/16#

(ii) Your equation is #y=-1/x^4+1/(2x)+45/16#