f(x)=x5−34x3+29x2+212x−300(a)upperboundsN=1+21300≈18171065111110171065−9<0−3425566229>0>041212>0>0458−300>0>0>0upperboundsf(x)=5
lowerboundsf(−x)=−x5+34x3+29x2−212x−300f1(x)=−f(−x)==x5−34x3−29x2+212x+300N=1+2134≈651105−34−9<0−29212300upperboundsf1(x)=6lowerboundsf(x)=−6x∈(−6,5)
(b)300:±1,±2,±3,±4,±5,±6,±10,±12,±15,±20,±25,±30,±50,±60,±75,±100,±150,±300x∈(−6,5)α:±1,±2,±3,±4,±5f(1)=1−34+29+212−300=−92f(−1)=−1+34+29−212−300=−450α−1f(1)∈Z,α+1f(−1)∈Z−α−192∈Z,−α−1450∈Zα=2,−32−311102−1−34−30−2729−3150212−1500−3000x1=−3,x2=2−rootsf(x)
(c)f(x):x+3x5−34x3+29x2+212x−300x5+3x4−−−−−−−−3x4−34x3+29x2+212x−300−3x4−9x3−−−−−−−−25x3+29x2+212x−300−25x3−75x2−−−−−−−104x2+212x−300104x2+312x−−−−−−−−100x−300−100x−300−−−−−−−0∣x+3x4−3x3−25x2−104x−100
(d)f(x)=x5−34x3+29x2+212x−300f′(x)=5x3−102x2+58x+212xn+1=xn−f′(xn)f(xn)xn=−3f(−3)=(−3)5−34(−3)3+29(−3)2++212(−3)−300=0f′(−3)=5(−3)4−102(−3)2+58(−3)+212=−475xn+1=−3−−4750=−3
Comments