Consider the given function f(x)=x2+2x+2 ‌ So, ‌ f(x)‌‌=x2+2x+1+1 ‌‌=(x+1)2+1 Here, f(x)∈[1,∞) and g(x)‌‌=−x2+2x−1 ‌‌=−(x2−2x+1) ‌‌=−(x−1)2 Here, g(x)∈(−∞,0] Then, ‌
f
g
(x)‌‌=‌
x2+2x+2
−x2+2x−1
‌‌=y So x2+2x+2‌‌=−yx2+2xy−y x2+yx2+2x−2xy+2+y‌‌=0 x2(1+y)+(2−2y)x+2+y‌‌=0 since, D≥0, so (2−2y)2−4(2+y)(1+y)≥0 4+4y2−8y−4(2+y)(1+y)&≥0 1+y2−2y−(2+y)(1+y)&≥0 1+y2−2y−2−3y−y2&≥0 Further simplify the above, −5y−1≥0 ‌‌‌‌y≤−‌