蟑螂 asked in 科學數學 · 1 decade ago

高一數(勘根定理)急~~10點

1.設f(x)=ax平-(a-1)x-6、a屬於實數,若方乘式f(x)=0有一根在1、2之間,另一根在-1、-2之間,求a之範圍?

2.當x三次-3x平-3x+k=0在1、2間有奇數個實根,求k之範圍?

3.(1)設f(x)=x三次-x-1,試證在1、2之間必有ㄧ實數c,使得f(c)=c。

(2)設f(x)=2x三次+7x平+3x-3,求證在0、1間必存在ㄧ實數p,使得f(p)=5p+1。

(3)設f(x)=x平+2x-5,g(x)=2x平-3,試證:(1)方程式f(x)+g(x)=0在1與2之間至少存在ㄧ實根。(2)方程式f(x)乘以g(x)=0在1與2之間至少存在ㄧ實根。

4.已知方程式x三方-8x+1=0在0與1之間有ㄧ個時根,求此根的近似值(正確至小數點後第ㄧ位)?

(以上皆要作法) 謝謝~~~~

1 Answer

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  • 1 decade ago
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    1.

    case 1: a>0(concave upward)

    f(2)=4a-2(a-1)-6>0, then a>2

    f(-2)=4a+2(a-1)-6>0, then a>4/3

    f(1)=a-(a-1)-6<0, then any a

    f(-1)=a+(a-1)-6<0, then a< 7/2

    so that, 2<a<7/2

    case 2: a<0(concave downward)

    f(2)=4a-2(a-1)-6<0, then a<2

    f(-2)= 4a+2(a-1)-6<0, then a<4/3

    f(1)=a-(a-1)-6 >0, then imposible

    Ans: 2<a<7/2

    2.

    f(x)=x^3-3x^2-3x+k

    f(1)*f(2)<0, then (k-5)(k-10)<0, so, 5<k<10

    3.

    (1)Consider the equation g(x)=f(x)-x=x^3-2x-1

    g(1)=-2, g(2)=3, then g(1)*g(2)<0

    By the intermediate value thm., there exists c in (1,2) and g(c)=0

    ie. 0=g(c)=f(c)-c, or f(c)=c

    (2)Consider the equation g(x)=f(x)-(5x+1)=2x^3+7x^2-2x-4

    g(0)= -4, g(1)=3, then g(0)g(1)<0, so, there exists p in (0,1) and g(p)=0

    ie. 0=g(p)=f(p)-(5p+1)=0, or f(p)=5p+1

    (3)

    (a) Let F(x)=f(x)+g(x)=3x^2+2x-8, then

    F(1)= -3, F(2)=8, F(1)*F(2)<0, so, there exists k in (1,2), such that F(k)=0

    ie. 0=F(k)=f(k)+g(k)

    (b) g(x)=2x^2 -3, g(1)*g(2)=(-1)*5<0, then

    g(x)=0 has one solution in (1, 2), so f(x)*g(x)=0 has at least one solution in (1,2)

    4. Let f(x)=x^3-8x+1, then

    f(0)*f(1)=1*(-6)<0,

    f(0.1)*f(0.2)=0.201*(-0.5~) <0

    f(0.1)*f(0.15)<0

    so that, the root in (0,1) approximates 0.1

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