Question

【題目】已知函數(shù)f(x)= x3-ax2,a∈R.

(1)當(dāng)a=2時,求曲線y=f(x)在點(3,f(3))處的切線方程;

(2)設(shè)函數(shù)g(x)=f(x)+(x-a)cos x-sin x,討論g(x)的單調(diào)性并判斷有無極值,有極值時求出極值.

Answer 1

【答案】（1）3x-y-9=0；（2）見解析.

【解析】試題分析：(1)求導(dǎo)，利用導(dǎo)數(shù)的幾何意義進(jìn)行求解；(2)求導(dǎo)，為了分析導(dǎo)函數(shù)的符號變化，再構(gòu)造新函數(shù)，利用導(dǎo)數(shù)研究新函數(shù)的單調(diào)性，進(jìn)而得到原函數(shù)的單調(diào)性和極值.

(1)由題意f'(x)=x2-ax,所以當(dāng)a=2時,f(3)=0,f'(x)=x2-2x,所以f'(3)=3,因此曲線y=f(x)在點(3,f(3))處的切線方程是y=3(x-3),即3x-y-9=0.

(2)因為g(x)=f(x)+(x-a)cos x-sin x,

所以g'(x)=f'(x)+cos x-(x-a)sin x-cos x

=x(x-a)-(x-a)sin x

=(x-a)(x-sin x).

令h(x)=x-sin x,則h'(x)=1-cos x≥0,所以h(x)在R上單調(diào)遞增.

因為h(0)=0,所以當(dāng)x>0時,h(x)>0;

當(dāng)x<0時,h(x)<0.

①當(dāng)a<0時,g'(x)=(x-a)(x-sin x),

當(dāng)x∈(-∞,a)時,x-a<0,g'(x)>0,g(x)單調(diào)遞增;

當(dāng)x∈(a,0)時,x-a>0,g'(x)<0,g(x)單調(diào)遞減;

當(dāng)x∈(0,+∞)時,x-a>0,g'(x)>0,g(x)單調(diào)遞增.

所以當(dāng)x=a時g(x)取到極大值,極大值是g(a)=- a3-sin a,

當(dāng)x=0時g(x)取到極小值,極小值是g(0)=-a.

②當(dāng)a=0時,g'(x)=x(x-sin x),當(dāng)x∈(-∞,+∞)時,g'(x)≥0,g(x)單調(diào)遞增;

所以g(x)在(-∞,+∞)上單調(diào)遞增,g(x)無極大值也無極小值.

③當(dāng)a>0時,g'(x)=(x-a)(x-sin x).

當(dāng)x∈(-∞,0)時,x-a<0,g'(x)>0,g(x)單調(diào)遞增;

當(dāng)x∈(0,a)時,x-a<0,g'(x)<0,g(x)單調(diào)遞減;

當(dāng)x∈(a,+∞)時,x-a>0,g'(x)>0,g(x)單調(diào)遞增.

所以當(dāng)x=0時g(x)取到極大值,極大值是g(0)=-a;

當(dāng)x=a時g(x)取到極小值,極小值是g(a)=- a3-sin a.

綜上所述:當(dāng)a<0時,函數(shù)g(x)在(-∞,a)和(0,+∞)上單調(diào)遞增,在(a,0)上單調(diào)遞減,函數(shù)既有極大值,又有極小值,極大值是g(a)=- a3-sin a,極小值是g(0)=-a;

當(dāng)a=0時,函數(shù)g(x)在(-∞,+∞)上單調(diào)遞增,無極值;

當(dāng)a>0時,函數(shù)g(x)在(-∞,0)和(a,+∞)上單調(diào)遞增,在(0,a)上單調(diào)遞減,函數(shù)既有極大值,又有極小值,極大值是g(0)=-a,極小值是g(a)=- a3-sin a.