求極值...微積分

Question

1.f(x)=cos^2xsin^4x，則f(x)的最大值為？

A：4/27

2.f(x)=x^4-4x^3+8x-3/x^2-2x+2，則當x=？時，f(x)有極小值？

A：1正負根號2，-2

3.設f(x)=│x│(x^2-3x)，
則f(x)在區間[-1,4]的最大值為？最小值為？

A：16，-4

4.設 f(x)=x^4+4x^3-20x^2-96x+70，
若f(x)在閉區間[-10,a]中是嚴格遞減，
而f(x)在區間[b,無限大)中是嚴格遞增，
則a的最大值為？b的最小值為？

Answer 1

1. f(x) = cos2x*sin4x = (1-sin2x)sin4x
設 sin(x) = A 則 f(A) = -A6 + A4
f' = -6A5 + 4A3
f' = 0 => A = 0 or A2 = 2/3
A = 0 => f(A) = 0
A2 = 2/3 => f(A) = -(2/3)3 + (2/3)2
= 4/27

2. f(x) = (x4 - 4x3 + 8x - 3)/(x2- 2x + 2)
= x2 - 2x - 6 + 9/(x2- 2x + 2)
f'(x) = 2x - 2 - 9(2x-2)/(x2- 2x + 2) = (2x-2)[1-9/(x2- 2x + 2)2]
f'(x) = 0 => x = 1 or (x2- 2x + 2)2 = 9
(x2- 2x + 2)2 = 9
=> x2- 2x + 2 = +- 3
若 x2- 2x + 2 = -3 => x2- 2x + 5 = 0 => 虛根不合
若 x2- 2x + 2 = 3 => x2- 2x - 1 = 0 => x = 1 +- √2
f''(x) = 2 - [18(x2- 2x + 2) - 72(x-1)2]/(x2- 2x + 2)3
f''(1) = -16 < 0 => 極大值
f''(1 +- √2) = 2 - [18(3) - 72(+- √2)2]/(3)3 = 16/3 > 0 => 極小值
f(x) = x2 - 2x - 6 + 9/(x2- 2x + 2)
= (x2 - 2x - 1) - 5 + 9/[(x2- 2x - 1) + 3]
f(1 +- √2) = 0 - 5 + 9/[0 + 3] = -5 + 3 = - 2
當 x= 1 +- √2時, f(x)有極小值 -2
3. 先不看 | | 則 f(x)= x(x2-3x) = x3 - 3x2
f'(x) = 3x2 - 6x
f'(x) = 0 => x = 0, 2
既然是求最大最小值 把極值及端點代入比較
剛才我們先忽略絕對值
就有可能把絕對值等於 0 的點的極性改變
所以在比較值的時候 該點也要代入
f(-1) = 4
f(0) = 0
f(2) = - 4 => 最小值
f(4) = 16 => 最大值
4. 設 f(x) = x4 + 4x3 - 20x2 - 96x + 70
f'(x) = 4x3 + 12x2 - 40x - 96
f'(x) = 0 => x3 + 3x2 - 10x - 24 => x = -4, -2, 3
因為 f(x) 是一個多項式函數 且 最高次項係數為正
所以 當 -4 < x < -2, or 3 < x 時 遞增
當 x < -4, or -2 < x < 3 時 遞減
所以 a 的最大值為 - 4, b的最小值為 3
如果有問題, 請來函討論. 不然, 我可能會錯失你再補充的疑點.