=y( let ) ⇒‌‌x4+x2+1=y(x2+1) and y>0 ⇒x4+(1−y)x2+(1−y)=0 ⇒x4+(1−y)x2+(1−y)=0 ∵x∈R, so discriminant ≥0 ⇒‌(1−y)2−4(1−y)≥0 ⇒‌(1−y)(1−y−4)≥0 ⇒‌(1−y)(−3−y)≥0 ⇒‌(y+3)(y−1)≥0 ⇒y∈‌(−∞,−3]∪[1,∞) but y>0, so y∈[1,∞) Therefore range of f is [1,∞)