Question

If x be real find the maximum and minimum value of: y=(x+2)/(2x2+3x+6)​

Answer 1

Step-by-step explanation:

y = x+2/2x²+3x+6     cross multiply

2x²y+3xy+6y-x-2=0     note that    x is real     so  D≥0

a=2y b=3y-1  and c = 6y-2     use D≥0

(3y-1)² - 4 × 2y ×(6y-2)≥0

9y²-6y+1-48y²+16y≥0

39y²-10y-1≤0

factorise   y = 10±√100+4×39/2×39

                 y = 10±16/78

                 y = 26/78  is maximum value  and y= -6/78 is minimum value

Answer 2

Answer:

maximum 1/3 and minimum -1/13