If x be real find the maximum and minimum value of: y=(x+2)/(2x2+3x+6)
Attachments:
Answers
Answered by
22
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
Answered by
0
Answer:
maximum 1/3 and minimum -1/13
Similar questions