find range function f(x)= x(√(7x−x^2 −1)) without calculus


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Question Number 83977 by john santu last updated on 08/Mar/20

$find range function$ $f (x) = x \sqrt{7 x - x^{2} - 1} without$ $calculus$
	Answered by john santu last updated on 08/Mar/20

	Answered by TANMAY PANACEA last updated on 08/Mar/20

	$x \times \sqrt{- x^{2} + 7 x - 1}$ $x \times \sqrt{- 1 - (x^{2} - 2 \times x \times \frac{7}{2} + \frac{49}{4} - \frac{49}{4})}$ $x \times \sqrt{- 1 + \frac{49}{4} - {(x - \frac{7}{2})}^{2}}$ $x \times \sqrt{\frac{45}{4} - {(x - \frac{7}{2})}^{2}}$ ${(x - \frac{7}{2})}^{2} ≯ \frac{45}{4}$ $\frac{45}{4} - {(x - \frac{7}{2})}^{2} ⩾ 0$ $(\frac{3 \sqrt{5}}{2} + x - \frac{7}{2}) (\frac{3 \sqrt{5}}{2} - x + \frac{7}{2}) ⩾ 0$ $c r i t i c a l v a l u e o f x = \frac{7 - 3 \sqrt{5}}{2} a n d \frac{7 + 3 \sqrt{5}}{2}$ $(x - a) (b - x) ⩾ 0 [a = \frac{7 - 3 \sqrt{5}}{2} b = \frac{7 + 3 \sqrt{5}}{2} b > a]$ $g (x) = (x - a) (b - x)$ $w h e n x > b g (x) < 0$ $w h e n x < a g (x) < 0$ $w h e n b > x > a g (x) > 0$ $g (x) = 0 w h e n x = a a n d x = b$ $f (x) = x \times \sqrt{- x^{2} + 7 x - 1}$ $f (x) = x \times \sqrt{(x - a) (b - x)} [b > a]$ $f (a) = f (b) = 0 b u t x ≯ b, x ≮ a$ $w a i t . . .$
	Commented by john santu last updated on 08/Mar/20

	$domain \frac{45}{4} - {(x - \frac{7}{2})}^{2} ⩾ 0$ ${(x - \frac{7}{2})}^{2} - {(\frac{3 \sqrt{5}}{2})}^{2} ⩽ 0$ $(x - \frac{7}{2} + \frac{3 \sqrt{5}}{2}) (x - \frac{7}{2} - \frac{3 \sqrt{5}}{2}) ⩽ 0$ $\frac{7 - 3 \sqrt{5}}{2} ⩽ x ⩽ \frac{7 + 3 \sqrt{5}}{2}$
	Commented by john santu last updated on 08/Mar/20

	$range y = \sqrt{- x^{4} + 7 x^{3} - x^{2}}$
	Commented by john santu last updated on 08/Mar/20

	$y^{'} = \frac{- 4 x^{3} + 21 x^{2} - 2 x}{2 \sqrt{- x^{4} + 7 x^{3} - x^{2}}} = 0$ $4 x^{3} - 21 x^{2} + 2 x = 0$ $x (4 x^{2} - 21 x + 2) = 0$ $x^{2} - \frac{21}{4} x + \frac{1}{2} = 0$ ${(x - \frac{21}{8})}^{2} + \frac{1}{2} - \frac{441}{64} = 0$
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