高一數(勘根定理)急~~10點

Question

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~~~~

Answer 1

Favorite Answer

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