导数求函数的单调性与极值习题

证明：当x>0x>0x>0时，sin⁡x+cos⁡x>−x2+x+1\sin x+\cos x>-x^2+x+1sinx+cosx>−x2+x+1

证明：

\qquad令f(x)=sin⁡x+cos⁡x+x2−x−1f(x)=\sin x+\cos x+x^2-x-1f(x)=sinx+cosx+x2−x−1

\qquad则f′(x)=cos⁡x−sin⁡x+2x−1f'(x)=\cos x-\sin x+2x-1f′(x)=cosx−sinx+2x−1

f′′(x)=−sin⁡x−cos⁡x+2=(1−sin⁡x)+(1−cos⁡x)\qquad f''(x)=-\sin x-\cos x+2=(1-\sin x)+(1-\cos x)f′′(x)=−sinx−cosx+2=(1−sinx)+(1−cosx)

\qquad当x>0x>0x>0时，f′′(x)≥0f''(x)\geq0f′′(x)≥0，即f′(x)f'(x)f′(x)单调递增，故f′(x)f'(x)f′(x)有最小值f′(0)=0f'(0)=0f′(0)=0

\qquad故f′(x)f'(x)f′(x)有最小值f′(0)=0f'(0)=0f′(0)=0，即恒有f(x)>f(0)=0f(x)>f(0)=0f(x)>f(0)=0

f(x)\qquad f(x)f(x)单调递增，即f(x)f(x)f(x)有最小值f(0)=0f(0)=0f(0)=0

x>0\qquad x>0x>0时f(x)>0f(x)>0f(x)>0，得证x>0x>0x>0时sin⁡x+cos⁡x>−x2+x+1\sin x+\cos x>-x^2+x+1sinx+cosx>−x2+x+1