Question

if X is real, find minimum of x^2-2x+2/x^2+3x+9​

Answer 1

""" ❤️ Answer ❤️ """

Let

y=

x 2

x 2

.

⇒ y(x

2

+ 2x+ 3)=

x 2

14x+ 9

⇒ (y− 1)x

2

+ 2(y− 7)x+ 3y− 9= 0

Since, x is real, so discriminant of above equation will be greater than or equal to 0.

D≥ 0

⇒ 4(y− 7)

2

− 4(y− 1)(3y− 9)≥ 0

⇒ (y− 7)

2

− (y− 1)(3y− 9)≥ 0

⇒

y 2

49− 14y− 3(y

2

− 4y+ 3)≥ 0

⇒ −2y

2

− 2y+ 40≥ 0

⇒

y 2

y− 20≤ 0

⇒

y 2

5y− 4y− 20≤ 0

⇒ y(y+ 5)− 4(y+ 5)≤ 0

⇒ (y+ 5)(y− 4)≤ 0

⇒ y∈ [−5,4]

So, the maximum value of y is 4 and minimum value of y is -4

Answer 2

Answer:

x²-2x+2/x²+3x+9

=9+x³+2/x²

-x=11

=x+11

Hope it helps

Brainliest