y=(x+4)+√(9-x²).换元,可设x=3cost,(0≤t≤π).则y=3cost+4+3sint=(3√2)sin[t+(π/4)]+4.即y=(3√2)sin[t+(π/4)]+4.0≤t≤π,===>π/4≤t+(π/4)≤5π/4.===>-√2/2≤sin[t+(π/4)]≤1.===>-3≤(3√2)sin[t+(π/4)]≤3√2.===>1≤y≤4+3√2.∴值域为[1,4+3√2].
y=x+4+根号下9-x^2值域为.
y=(x+4)+√(9-x²).换元,可设x=3cost,(0≤t≤π).则y=3cost+4+3sint=(3√2)sin[t+(π/4)]+4.即y=(3√2)sin[t+(π/4)]+4.0≤t≤π,===>π/4≤t+(π/4)≤5π/4.===>-√2/2≤sin[t+(π/4)]≤1.===>-3≤(3√2)sin[t+(π/4)]≤3√2.===>1≤y≤4+3√2.∴值域为[1,4+3√2].