Question

已知函数f(x)=ax^3-3/2x^2+1(x属于r),其中a>0 若a=1，求曲线y=f（x）在点（2，f（2））处的切线方程.

若a>0,b=2,当x∈[-1,1]时,求f(x)的最小值

Answer 1

f(x) = ax^3 - (3/2)x^2 + 1，x∈R，其中a > 0
当a = 1
f(x) = x^3 - (3/2)x^2 + 1，f(2) = 2^3 - (3/2)(2)^2 + 1 = 3
f'(x) = 3x^2 - 3x
f'(2) = 3(2)^2 - 3(2) = 6
切线方程：y - 3 = 6(x - 2) ==> y = 6x - 9

若a > 0，b = 2，x∈[- 1，1]，求f(x)的最小值？
b在哪里？

追问

已知函数f(x)=ax^3-3/2x^2+b(x属于r),，(1)曲线y=f（x）在点（2，f（2））处的切线方程y=6x-8,求a、b的值. (2)其中a>0 b=2，当x∈[-1,1]时,求f(x)的最小值

追答

f(x) = ax^3 - (3/2)x^2 + b，x∈R
f'(x) = 3ax^2 - 3x
切线y = 6x - 8的斜率是6
所以f'(2) = 6
3a(2)^2 - 3(2) = 6
a = 1
将点(2，f(2))代入y = 6x - 8
f(2) = 6(2) - 8 = 4
f(x) = x^3 - (3/2)x^2 + b
f(2) = 4 ==> 2^3 - (3/2)(2)^2 + b = 4 ==> b = 2

a > 0，b = 2，当x∈[- 1，1]，求f(x)的最小值？，a的值没给？
f(x) = ax^3 - (3/2)x^2 + 2
f'(x) = 3ax^2 - 3x = 0
f''(x) = 6ax - 3
x(ax - 1) = 0
x = 0 或 x = 1/a
f''(0) = - 3 0，取得极小值
极小值f(1/a) = 2 - 1/(2a^2) ≤ 2
f(- 1) = - a + 1/2 ≤ 1/2
f(1) = a + 1/2 ≥ 1/2
所以f(x)的最小值是- a + 1/2

Answer 2

b在哪里？