[10pt]polynomial

3x^4+2x^2+3x+1
=3(x^2-1/6)^2+3(x+1/2)^2+1/6
>0

人地講到明兩幅圖都係 y = 3 x^4 + 2 x^2 + 3 x + 1...
只不過 一幅由 -1 到 +1, 一幅 -10 到 +10, 你答左 D 咩出黎ar????

2008-07-31 12:35:19 補充：
Algebraical approach is as follows:

Firstly, we complete square and somehow make the troublesome 3x disappear.
y = 3 x^4 + 2 x^2 + 3 x + 1
= 3 x^4 + (rt2 x)^2 + 2* (rt2 x)* (3/ 2rt2) + (3/2rt2)^2 - (3/2rt2)^2 + 1
= 3 x^4 + ( rt2 x + 3/ 2rt2 )^2 - 1/8
Note that, 3 x^4 and (rt2+3/2rt2)^2 are always non-negative.

our target becomes to show that 3 x^4 + ( rt2 x + 3/ 2rt2 )^2 always strictly greater than 1/8
3 x^4 is strictly greater than 1/8 outside +-(1/24)^(1/4) which are -0.4518 and 0.4518
( rt2 x + 3/ 2rt2 )^2 is strictly greater than 1/8 outside -3/4 +- 1/4 , which are -1 and -1/2
Therefore, at any value of x, 3 x^4 + ( rt2 x + 3/ 2rt2 )^2 is strictly greater than 1/8.

Thus, y is always positive!

用Quartic Solver，冇real root，做完。

2008-07-26 22:57:11 補充：
http://www.akiti.ca/Quad4Deg.html

Thank you for providing the curves of f(x) = 3x4 and
f(x) = 2x2 + 3x + 1
As follow,
y = 3x4 + 2x2 + 3x + 1
can be divided into two functions
f(x) = 3x4 and f(x) = -(2x2 + 3x + 1)
如果f(x) = 3x4 and f(x) = -(2x2 + 3x + 1)沒有交點
則
y = 3x4 + 2x2 + 3x + 1 =/= 0
From the curves above
the second curve should folded below the x- axis in order to get
the curve of f(x) = -( 2x2 + 3x + 1 )
可見
f(x) = 3x4 and f(x) = -(2x2 + 3x + 1)係沒有交點
因此
y = 3x4 + 2x2 + 3x + 1 =/= 0