數學資料庫 (http://www.mathdb.org/ 及 http://eng.mathdb.org/)是香港其中一個最大型的數學網站，由一班來自香港各間大學及美國多間大學的數學系、物理系、精算系、建築系、電子工程系的學生建立。數學資料庫旨在於網上提供豐富的數學資源。我們上載了不同種類和數學相關的教學資料，是同學們一個好的參考網站。
證明方程 x4 - 8x3 + 22x2 - 24x + 10 = 0 沒有實根。
Let f(x) be the given functioncheck f(0) = 10, f(1) = 1, f(2) = 2so, f(0),f(1),f(2) <> 0 (mode 3)Thus, there is no integral roots for f(x).However, the question asks for proving no real roots.========================If x < 0, f(x) > 0. thus no real roots for x <0.f'(x) = 4x^3 - 24x^2 + 44x - 24set f'(x) = 0=> x^3 - 6x^2 + 11x - 6 = 0=> (x-1)(x-2)(x-3) = 0so the three turning points are at x=1, 2, 3we have f(0) = 10, f(1) = 1, f(2) = 2check f(3) = 1and f(x>3) > 0 as f'(x>3) > 0It is not difficult to sketch the graph of f(x)Thus f(x) > 0 for all real x.
Here is a simpler proof: f(x) = (x^2 - 4x + 3)^2 + 1. Done.
哈，f(x) = (x^2 - 4x + 3)^2 + 1 是 Kahoo 的「立題原意」嗎？
路过~~~
不，「立題原意」是用微分的，不過 Koopa 的做法令我想到一些出題的好主意……
ha, actually I tried to completing the square...but not get it done...so give up. :D
我的做法和POP差不多:differentiate f(x)=0,=> 4(x-1)(x-2)(x-3)=0=> x=1,2,3sub into f(x),f(1)=1f(2)=2f(3)=1but f(x)=0so it is in consistentso f(x)have no real solution
Let f(x) be the given function
回覆刪除check
f(0) = 10, f(1) = 1, f(2) = 2
so, f(0),f(1),f(2) <> 0 (mode 3)
Thus, there is no integral roots for f(x).
However, the question asks for proving no real roots.
========================
If x < 0, f(x) > 0. thus no real roots for x <0.
f'(x) = 4x^3 - 24x^2 + 44x - 24
set f'(x) = 0
=> x^3 - 6x^2 + 11x - 6 = 0
=> (x-1)(x-2)(x-3) = 0
so the three turning points are at x=1, 2, 3
we have
f(0) = 10, f(1) = 1, f(2) = 2
check f(3) = 1
and f(x>3) > 0 as f'(x>3) > 0
It is not difficult to sketch the graph of f(x)
Thus f(x) > 0 for all real x.
Here is a simpler proof:
回覆刪除f(x) = (x^2 - 4x + 3)^2 + 1. Done.
哈，f(x) = (x^2 - 4x + 3)^2 + 1 是 Kahoo 的「立題原意」嗎？
回覆刪除路过~~~
回覆刪除不，「立題原意」是用微分的，不過 Koopa 的做法令我想到一些出題的好主意……
回覆刪除ha, actually I tried to completing the square...but not get it done...so give up. :D
回覆刪除我的做法和POP差不多:
回覆刪除differentiate f(x)=0,
=> 4(x-1)(x-2)(x-3)=0
=> x=1,2,3
sub into f(x),
f(1)=1
f(2)=2
f(3)=1
but f(x)=0
so it is in consistent
so f(x)have no real solution