Sample Solution from
Numerical Analysis
3rd Edition
ISBN: 9780134696454
Chapter 0.5
Problem 1E
Try another sample solution

a.

To determine

To prove: f(c)=0 for some 0<c<1 : f(x)=x34x+1 .

Expert Solution

Explanation of Solution

Given information: f(x)=x34x+1

Theorem used:

Intermediate Value Theorem: If f is a continuous function on [a,b] and if y is a number between f(a) and f(b), then there exists a number c with acb such that f(c)=y .

Calculation:

Consider the equation, f(x)=x34x+1

Substituting x=1 in f(x)=x34x+1 , we get

  f(1)=134(1)+1=14+1=2

Substituting x=0 in f(x)=x34x+1 , we get

  f(0)=034(0)+1=00+1=1

Therefore, we have

  2<0<1f(1)<0<f(0)

Using Intermediate value Theorem, there exists c(0,1) such that f(c)=0 .

Hence, it is proved.

b.

To determine

To prove: f(c)=0 for some 0<c<1 : f(x)=5cosπx4 .

Expert Solution

Explanation of Solution

Given information: f(x)=5cosπx4

Theorem used:

Intermediate Value Theorem: If f is a continuous function on [a,b] and if y is a number between f(a) and f(b), then there exists a number c with acb such that f(c)=y .

Calculation:

Consider the equation, f(x)=5cosπx4

Substituting x=1 in f(x)=5cosπx4 , we get

  f(1)=5cosπ4=5(1)4=54=9

Substituting x=0 in f(x)=5cosπx4 , we get

  f(0)=5cosπ(0)4=54=1

Therefore, we have

  9<0<1f(1)<0<f(0)

Using Intermediate value Theorem, there exists c(0,1) such that f(c)=0

Hence, it is proved.

c.

To determine

To prove: f(c)=0 for some 0<c<1 : f(x)=8x48x2+1 .

Expert Solution

Explanation of Solution

Given information: f(x)=8x48x2+1

Theorem used:

Intermediate Value Theorem: If f is a continuous function on [a,b] and if y is a number between f(a) and f(b), then there exists a number c with acb such that f(c)=y .

Calculation:

Consider the equation, f(x)=8x48x2+1

Substituting x=1 in f(x)=8x48x2+1 , we get

  f(1)=8(1)48(1)2+1=88+1=1

Substituting x=0 in f(x)=8x48x2+1 , we get

  f(0)=8(0)48(0)2+1=00+1=1

Substituting x = 12 in f(x)=8x48x2+1, we get

  f(12)=8( 1 2)48( 1 2)2+1=8(1 16)8(14)+1=122+1=12

Therefore, we have

  12<0<1f(12)<0<f(0)

Using Intermediate value Theorem, there exists c(0,12) such that f(c)=0 .

Hence, it is proved.

Not sold yet?Try another sample solution