Question

已知函數(shù)f(x)=x3-x2+x+b,其中a,b∈R.

(1)若曲線y=f(x)在點P(2,f(2))處的切線方程為y=5x-4,求函數(shù)f(x)的解析式.

(2)當(dāng)a>0時,討論函數(shù)f(x)的單調(diào)性.

 

Answer 1

(1) f(x)=x3-2x2+x+4

(2) 當(dāng)0<a<1時,>1,函數(shù)f(x)在區(qū)間(-∞,1)及(,+∞)上為增函數(shù),在區(qū)間(1,)上為減函數(shù);

當(dāng)a=1時,=1,函數(shù)f(x)在區(qū)間(-∞,+∞)上為增函數(shù);

當(dāng)a>1時,<1,函數(shù)f(x)在區(qū)間(-∞,)及(1,+∞)上為增函數(shù),在區(qū)間(,1)上為減函數(shù).

【解析】(1)f'(x)=ax2-(a+1)x+1.

由導(dǎo)數(shù)的幾何意義得f'(2)=5,于是a=3.

由切點P(2,f(2))在直線y=5x-4上可知2+b=6,解得b=4.

所以函數(shù)f(x)的解析式為f(x)=x3-2x2+x+4.

(2)f'(x)=ax2-(a+1)x+1=a(x-)(x-1).

當(dāng)0<a<1時,>1,函數(shù)f(x)在區(qū)間(-∞,1)及(,+∞)上為增函數(shù),在區(qū)間(1,)上為減函數(shù);

當(dāng)a=1時,=1,函數(shù)f(x)在區(qū)間(-∞,+∞)上為增函數(shù);

當(dāng)a>1時,<1,函數(shù)f(x)在區(qū)間(-∞,)及(1,+∞)上為增函數(shù),在區(qū)間(,1)上為減函數(shù).