设x=sint,t∈[-π/2,π/2],由此cost≥0;
f(x)变形为f(t)=4-2sin²t+sint√(1-sin²t)
=4-2sin²t+sintcost
=3+cos2t+0.5sin2t
=3+sin(2t+φ)√(1²+0.5²) 注:sinφ=1/√(1²+0.5²) cosφ=0.5/√(1²+0.5²)
由于sin(2t+φ)介于±1,所以f(t)介于3±√(1²+0.5²) 之间.
设x=sint,t∈[-π/2,π/2],由此cost≥0;
f(x)变形为f(t)=4-2sin²t+sint√(1-sin²t)
=4-2sin²t+sintcost
=3+cos2t+0.5sin2t
=3+sin(2t+φ)√(1²+0.5²) 注:sinφ=1/√(1²+0.5²) cosφ=0.5/√(1²+0.5²)
由于sin(2t+φ)介于±1,所以f(t)介于3±√(1²+0.5²) 之间.