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Question Number 58971 by hovea cw last updated on 02/May/19

Commented by hovea cw last updated on 02/May/19

$hd plz$
	Answered by tanmay last updated on 02/May/19

	$y = x^{3} + {p x}^{2} + q x + r$ $\frac{d y}{d x} = 3 x^{2} + 2 p x + q$ $\frac{d^{2} y}{{d x}^{2}} = 6 x + 2 p$ $l e t p o i n t N (α, 0) a t p o i n t N y_{m i n i m u m}$ $1) \frac{d y}{d x} = 0 \to 3 α^{2} + 2 p α + q = 0 f o r m i n$ $2) \frac{d^{2} y}{{d x}^{2}} > 0 f o r m i n i m u m$ $3 α^{2} + 2 p α + q = 0$ $α = \frac{- 2 p \pm \sqrt{4 p^{2} - 12 q}}{6}$ $α = \frac{- p \pm \sqrt{p^{2} - 3 q}}{3}$ $f r o m g i v e n f i g u r e i t i s c l e a r α > 0$ $p o i n t N (\frac{- p + \sqrt{p^{2} - 3 q}}{3}, 0)$ ${(\frac{d^{2} y}{{d x}^{2}})}_{x = α} = 6 α + 2 p$ $6 (\frac{- p + \sqrt{p^{2} - 3 q}}{3}) + 2 p > 0$ $2 (\sqrt{p^{2} - 3 q}) > 0 [p^{2} > 3 q]$ $n o w f o r p o i n t K (0, β)$ $y = x^{3} + {p x}^{2} + q x + r$ $β = r$ $\frac{d y}{d x} = 3 x^{2} + 2 p x + q$ ${(\frac{d y}{d x})}_{x = 0} 3 \times 0^{2} + 2 \times p \times 0 + q = 0 [f o r m a x \frac{d y}{d x} = 0]$ $q = 0$ ${(\frac{d^{2} y}{{d x}^{2}})}_{x = 0} 6 \times 0 + 2 p < 0 p < 0$ $p < 0$ $q = 0$ $r = β \to K (0, β) β \leftarrow o r d i n a t e o f p o i n t K$ $i h a v e t r i e d t o u n d e r s t a n d t h e q u e s t i o n . . .$
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