三次多項式圖形探討

• 1. () 12 x 3 +ax2 + bx+ c x x- a x 3x 3 + px+ q f (x = x3 + px+ q)p>0 f (x = x3 - px+ q p>0) x 3 + px+ q g (x = x3 + pxgf )gx g (x = x3 + px ) x b > a g ( )- g( )= b 3 + p - a 3 - p = ( - a )( 2 + ba + a 2 + p ba b a bb )b 2 +ba + a 2 + p g( b)- g( )> 0 g (x = x3 + px a)x 3 + px2 x 3 + px+ qx f(x)=0 () 1

2. 3 3p>0g ( x)= x 3 - px y=x y=pxy=x ()y=px(p>0)()g(x)x0() g (x = x3 - pxx>0 ) 2 3. x>0 x=a x>a g ( x ) - g (a ) = x 3 - a 3 - p ( x - a ) = ( x - a )( x 2 + xa + a 2 - p ) 0 0a x 2 + x + a 2 - p 0 a 00a = p>>0g 2 +gb + b 2 > 3a 2 = p g(g)- g( )> 0 b00q = pf(x)3 3() 2pf(x)=0 p q = - pf(x) 33 3()2 p 2 pf(x)=0- pq-p pf(x) 3 3 3 3 34 5. () 2p2 pf(x)=0q >pq < - p f(x) 3 3 3 3 2pq >p 3 3 ()f(x)=05 6. f(x) f(x)=0 qp3f(x)=x f(x)=03f(x)=x +q,q0f(x)=0 3 f(x)=x +px+q,p>0f(x)=03 f(x)=x px+q,p>0 f(x)=0 2 3 2p27q =4pq = p 3 33f(x)=x px+q,p>0f(x)=02 32 p2p27q 02p q > p 2 33 3 f(x)=0 27q >4p 2pq < - p 3 3 p=0 2 1 3 b + ba + a = b + a + a 2 0 2 2 2 4 p>0f(x)=0 - p p 3 3 http://scicomp.math.ntu.edu.tw/calculus/ Sturm 6