find-the-turning-point-on-the-curve-y-16-x-x-3-3-and-determine-wether-it-is-a-point-of-maximum-or-minimum-

Question Number 7401 by txsnims last updated on 27/Aug/16

f i n d t h e t u r n i n g p o i n t o n t h e c u r v e y = \frac{16}{x} + \frac{x^{3}}{3} a n d d e t e r m i n e w e t h e r i t i s a p o i n t o f m a x i m u m o r m i n i m u m

Answered by FilupSmith last updated on 27/Aug/16

y = 16 x^{- 1} + 3^{- 1} x^{3}

turning point at y^{'} = 0

y^{^{'}} = - 16 x^{- 2} + x^{2} = 0

x^{2} = \frac{16}{x^{2}}

x^{4} = 16

{(x^{2})}^{2} = {(\pm 4)}^{2}

x = \pm 2 for x \in R

x = + 2

y^{″} = 32 x^{- 3} + 2 x

y^{″} = 32 (\frac{1}{8}) + 4 (2)

y^{″} > 0

∴ minimum when x = 2

x = - 2

y^{″} = 32 (\frac{1}{- 8}) + 4 (- 2)

y^{″} < 0

∴ maximum when x = - 2