作业帮导数题,求详细解答
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导数题,求详细解答
导数题,求详细解答
导数题,求详细解答
f(x)=x³+bx²+cx+d过(0,2),得d=2
f'(x)=3x²+2bx+c
M(-1,f(-1))为切点,得 f'(-1)=3-2b+c=6 (切线斜率=6)即-2b+c=3
f(-1)=-1+b-c+2=-6+7=1 (M点也在切线y=6x+7上),即b-c=0
由上两式联立,得b=-3,c=-3
∴f(x)=x³-3x²-3x+2
f'(x)=3x²-6x-3=3(x²-2x-1)=3(x-1)²-6
f''(x)=6x-6=6(x-1)
f'(x)=0时,x=1±√2
f''(1+√2)>0,此点有极小值
f''(1-√2)
ƒ(x) = x³ + bx² + cx + d
图像过点P(0，2) ==> ƒ(0) = 2 ==> d = 2
将点M代入直线方程中得：6(- 1) - ƒ(- 1) + 7 = 0 ==> ƒ(- 1) = 1
将点M代入曲线方程得：ƒ(- 1) = - 1 + b - c + 2 = b - c...
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ƒ(x) = x³ + bx² + cx + d
图像过点P(0，2) ==> ƒ(0) = 2 ==> d = 2
将点M代入直线方程中得：6(- 1) - ƒ(- 1) + 7 = 0 ==> ƒ(- 1) = 1
将点M代入曲线方程得：ƒ(- 1) = - 1 + b - c + 2 = b - c + 1
即b - c + 1 = 1 ==> b = c
ƒ'(x) = 3x² + 2bx + c
切线方程的斜率 = - A/B = - 6/(- 1) = 6
在x = - 1处的斜率ƒ'(- 1) = 3 - 2b + c
即3 - 2b + c = 6 ==> c - 2b = 3
b - 2b = 3 ==> - b = 3 ==> b = - 3 = c
所以ƒ(x) = x³ - 3x² - 3x + 2
ƒ'(x) = 3x² - 6x - 3
令ƒ'(x) = 0 ==> 3x² - 6x - 3 = 0 ==> x = 1 - √2 或 x = 1 + √2
ƒ''(x) = 6x - 6
ƒ''(1 - √2) = - 6√2 < 0，取得极大值ƒ(1 - √2)
ƒ''(1 + √2) = 6√2 > 0，取得极小值ƒ(1 + √2)
当x < 1 - √2时ƒ'(x) > 0，递增
当1 - √2 < x < 1 + √2时ƒ'(x) < 0，递减
当x > 1 + √2时ƒ'(x) > 0，递增
∴递增区间：(- ∞，1 - √2]U[1 + √2，+∞)
∴递减区间：[1 - √2，1 + √2]
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