三次函數(shù)切線題目及答案一、選擇題（共30分）1.（3分）若函數(shù)f(x)=x^3-3x+1的切線斜率為0，則切點(diǎn)的橫坐標(biāo)為：A.1B.3C.-1D.0答案：A2.（3分）若函數(shù)f(x)=x^3-6x^2+11x-6的切線斜率為0，則切點(diǎn)的橫坐標(biāo)為：A.1B.2C.3D.4答案：B3.（3分）若函數(shù)f(x)=x^3-3x^2+2x+1的切線斜率為0，則切點(diǎn)的橫坐標(biāo)為：A.1B.2C.-1D.0答案：C4.（3分）若函數(shù)f(x)=x^3-6x^2+11x-6的切線斜率為k，則切點(diǎn)的橫坐標(biāo)為：A.kB.2kC.3kD.4k答案：B5.（3分）若函數(shù)f(x)=x^3-3x^2+2x+1的切線斜率為k，則切點(diǎn)的橫坐標(biāo)為：A.kB.2kC.-kD.0答案：C6.（3分）若函數(shù)f(x)=x^3-6x^2+11x-6的切線斜率為k，則切點(diǎn)的橫坐標(biāo)為：A.kB.2kC.3kD.4k答案：C7.（3分）若函數(shù)f(x)=x^3-3x^2+2x+1的切線斜率為k，則切點(diǎn)的縱坐標(biāo)為：A.k^3-3k^2+2k+1B.k^3-3k^2+2k-1C.k^3-3k^2+2k+2D.k^3-3k^2+2k-2答案：A8.（3分）若函數(shù)f(x)=x^3-6x^2+11x-6的切線斜率為k，則切點(diǎn)的縱坐標(biāo)為：A.k^3-6k^2+11k-6B.k^3-6k^2+11k+6C.k^3-6k^2+11k-12D.k^3-6k^2+11k+12答案：A9.（3分）若函數(shù)f(x)=x^3-3x^2+2x+1的切線斜率為k，則切點(diǎn)的縱坐標(biāo)為：A.k^3-3k^2+2k+1B.k^3-3k^2+2k-1C.k^3-3k^2+2k+2D.k^3-3k^2+2k-2答案：A10.（3分）若函數(shù)f(x)=x^3-6x^2+11x-6的切線斜率為k，則切點(diǎn)的縱坐標(biāo)為：A.k^3-6k^2+11k-6B.k^3-6k^2+11k+6C.k^3-6k^2+11k-12D.k^3-6k^2+11k+12答案：A二、填空題（共30分）1.（3分）函數(shù)f(x)=x^3-3x+1的導(dǎo)數(shù)為f'(x)=________。答案：3x^2-32.（3分）函數(shù)f(x)=x^3-6x^2+11x-6的導(dǎo)數(shù)為f'(x)=________。答案：3x^2-12x+113.（3分）函數(shù)f(x)=x^3-3x^2+2x+1的導(dǎo)數(shù)為f'(x)=________。答案：3x^2-6x+24.（3分）函數(shù)f(x)=x^3-6x^2+11x-6的二階導(dǎo)數(shù)為f''(x)=________。答案：6x-125.（3分）函數(shù)f(x)=x^3-3x^2+2x+1的二階導(dǎo)數(shù)為f''(x)=________。答案：6x-66.（3分）函數(shù)f(x)=x^3-3x+1的三階導(dǎo)數(shù)為f'''(x)=________。答案：67.（3分）函數(shù)f(x)=x^3-6x^2+11x-6的三階導(dǎo)數(shù)為f'''(x)=________。答案：68.（3分）函數(shù)f(x)=x^3-3x^2+2x+1的三階導(dǎo)數(shù)為f'''(x)=________。答案：69.（3分）函數(shù)f(x)=x^3-6x^2+11x-6的四階導(dǎo)數(shù)為f''''(x)=________。答案：010.（3分）函數(shù)f(x)=x^3-3x^2+2x+1的四階導(dǎo)數(shù)為f''''(x)=________。答案：0三、計(jì)算題（共20分）1.（10分）求函數(shù)f(x)=x^3-3x^2+2x+1在x=1處的切線方程。解：首先求導(dǎo)數(shù)f'(x)=3x^2-6x+2，代入x=1得到切線斜率k=-1。然后求切點(diǎn)坐標(biāo)(1,-1)。最后利用點(diǎn)斜式方程得到切線方程為y+1=-(x-1)，即x+y=0。2.（10分）求函數(shù)f(x)=x^3-6x^2+11x-6在x=2處的切線方程。解：首先求導(dǎo)數(shù)f'(x)=3x^2-12x+11，代入x=2得到切線斜率k=-1。然后求切點(diǎn)坐標(biāo)(2,1)。最后利用點(diǎn)斜式方程得到切線方程為y-1=-(x-2)，即x+y-3=0。四、證明題（共20分）1.（10分）證明函數(shù)f(x)=x^3-3x^2+2x+1在x=1處的切線斜率為-1。證明：首先求導(dǎo)數(shù)f'(x)=3x^2-6x+2，代入x=1得到f'(1)=-1。因此，函數(shù)f(x)=x^3-3x^2+2x+1在x=1處的切線斜率為-1。2.（10分）證明函數(shù)f(x)=x^3-6x^2+11