find minimun and maksimum y=((x^3 −1)/(x^2 −3))


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Question Number 121368 by abdullahquwatan last updated on 07/Nov/20

$find minimun and maksimum$ $y = \frac{x^{3} - 1}{x^{2} - 3}$
	Answered by MJS_new last updated on 07/Nov/20
	$x^3 −1=0 ⇒ zero at x=1 x^2 −3≠0 ⇒ asymptotes x=±(√3) ⇒ −∞<y<+∞ ((x^3 −1)/(x^2 −3))=x+((3x−1)/(x^2 −3)) ⇒ asymptote y=x y′=((x(x^3 −9x+2))/((x^2 −3)^2 )) y′′=((6(x^3 −x^2 +9x−1))/((x^2 −3)^3 )) solving y′=0 and testing y′′ { ((<0 (maximum))),((=0 (flat point))),((>0 (minimum))) :} leads to x_1 ≈−3.10548 local max with y≈−4.65822 x_2 =0 local minimum with y=(1/3) x_3 ≈.223462 local maximum with y≈.335193 x_4 ≈2.88202 local minimum with y≈4.32303$
	$x^{3} - 1 = 0 \Rightarrow zero at x = 1$ $x^{2} - 3 \neq 0 \Rightarrow asymptotes x = \pm \sqrt{3}$ $\Rightarrow - \infty < y < + \infty$ $\frac{x^{3} - 1}{x^{2} - 3} = x + \frac{3 x - 1}{x^{2} - 3} \Rightarrow asymptote y = x$ $y^{'} = \frac{x (x^{3} - 9 x + 2)}{{(x^{2} - 3)}^{2}}$ $y^{″} = \frac{6 (x^{3} - x^{2} + 9 x - 1)}{{(x^{2} - 3)}^{3}}$ $solving y^{'} = 0 and testing y^{″} {\begin{cases} < 0 (maximum) \\ = 0 (flat point) \\ > 0 (minimum) \end{cases}$ $leads to$ $x_{1} \approx - 3.10548 local \max with y \approx - 4.65822$ $x_{2} = 0 local minimum with y = \frac{1}{3}$ $x_{3} \approx . 223462 local maximum with y \approx . 335193$ $x_{4} \approx 2.88202 local minimum with y \approx 4.32303$
	Commented by abdullahquwatan last updated on 07/Nov/20

	$thank you sir$
	Commented by MJS_new last updated on 07/Nov/20

	$you are welcome$
	Answered by TANMAY PANACEA last updated on 07/Nov/20

	$\frac{d y}{d x} = \frac{(x^{2} - 3) (3 x^{2}) - (x^{3} - 1) 2 x}{{(x^{2} - 3)}^{2}}$ $\frac{d y}{d x} = \frac{3 x^{4} - 9 x^{2} - 2 x^{4} + 2 x}{{(x^{2} - 3)}^{2}} = \frac{x^{4} - 9 x^{2} + 2 x}{{(x^{2} - 3)}^{2}}$ $f o r m a x / m i n \frac{d y}{d x} = 0 x (x^{3} - 9 x + 2) = 0$ $x = 0 a n d 0.223, 2.882$ $w h e n x > 2.882 \frac{d y}{d x} > 0$ $w h e n 2.882 > x > 0.223 \frac{d y}{d x} < 0$ $w h e n 0.223 > x > 0 \frac{d y}{d x} > 0$ $w h e n x < 0 \frac{d y}{d x} < 0$ $s i g n o f \frac{d y}{d x} c h a n g e s f r o m - v e t o + v e w h e n c u r v e c r o s s x = 0$ $s o a t x = 0 g i v e n f u n c t i o n m i n i m u m$ $y_{x = 0} = \frac{1}{3} (m i n i m u m) = \frac{1}{3}$ $s i g n c h a n g e f r o m - v e t o + v e w h e n c r o s s x = 0.223$ $s o m a x i m u m y_{x = 0.223} = \frac{{(0.223)}^{3} - 1}{{(0.223)}^{2} - 3} (m a x i m u m) =$ $w h e n 2.882 > x \frac{d y}{d x} < 0$ $w h e n x > 2.882 \frac{d y}{d x} > 0$ $s i g n c h a n g e f r o m - v e t o + v e$ $s o m i n i m u m$ $y_{x = 2.882} = \frac{{(2.882)}^{3} - 1}{{(2.882)}^{2} - 3} (m i n i m u m) =$
	Commented by TANMAY PANACEA last updated on 07/Nov/20

	$p l s c a l c u l a t e t h e [v a l u e$
	Commented by abdullahquwatan last updated on 07/Nov/20

	$thank you sir$
	Commented by TANMAY PANACEA last updated on 07/Nov/20

	$m o s t w e l c o m e$
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