Question

If x is real the minimum and maximum value of x^2+14x+9/x^2+2x+3. Is

Answer 1

EXPLANATION.

→ x is real the minimum and maximum value of

x² + 14x + 9 / x² + 2x + 3.

→ Let y = x² + 14x + 9 / x² + 2x + 3

→ y ( x² + 2x + 3 ) = x² + 14x + 9

→ yx² + 2xy + 3y = x² + 14x + 9

→ ( y - 1 )x² + 2 ( y - 7 )x + 3y - 9 = 0

→ if x is real, → D ≥ 0

→ D = b² - 4ac ≥ 0

→ 4 ( y - 7 )² - 4( y - 1 ) ( 3y - 9 ) ≥ 0

→ 4 ( y² + 49 - 14y ) - 4 ( 3y² - 9y - 3y + 9 ) ≥ 0

→ y² + 49 - 14y - 3y² + 9y + 3y - 9 ≥ 0

→ - 2y² - 2y + 40 ≥ 0

→ y² + y - 20 ≤ 0

→ y² + 5y - 4y - 20 ≤ 0

→ y ( y + 5 ) - 4 ( y + 5 ) ≤ 0

→ ( y - 4 ) ( y + 5 ) ≤ 0

→ put the point on wavy curve and plot them.

→ we can put the number greater than 4 as

we can see it comes positive.

→ y € [ - 5 , 4 ]

→ maximum value = 4

→ minimum value = -5

Answer 2

$\: y = \frac{ {x}^{2} + 4x + 9}{( {x}^{2} + 2 x + 3) }$

${x}^{2} y + 2xy + 3y + {x}^{2} + 14x + 9$

${x}^{2} (y - 1) - 2x(y - 7) + 3(y - 3) = 0$

$\red{ \cancel4 \: (y - 7 {)}^{2}\cancel4(y - 1)3(y - 3) = 0 }$

$\red{ {y}^{2} + 49 - 14 y - ( {y}^{2} - 4y + 3)}$

$\red{ {y}^{2} + 49 - 14y - 3 {y}^{2} + 12y - 9 = 0 }$

$\red{ { - 2y}^{2} - 2y + 40 = 0}$

$\red{ {y}^{2} + y - 20 = 0}$

$\red{ {y}^{2} + 5y - 4y - 2 = 0}$

$\pink{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: (y + 5)(y - 4) = 0}$


$\pink{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: y = - 5,4}$