To prove : f ( c ) = 0 for some 0 < c < 1 : f ( x ) = x 3 − 4 x + 1 .

Question

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Answer 1

Question

Chapter 0.5, Problem 1E

a.

To determine

To prove: f(c)=0 for some 0<c<1 : f(x)=x3−4x+1 .

a.

Expert Solution

Explanation of Solution

Given information: f(x)=x3−4x+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 a≤c≤b such that f(c)=y .

Calculation:

Consider the equation, f(x)=x3−4x+1

Substituting x=1 in f(x)=x3−4x+1 , we get

f(1)=13−4(1)+1=1−4+1=−2

Substituting x=0 in f(x)=x3−4x+1 , we get

f(0)=03−4(0)+1=0−0+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)=5cosπx−4 .

b.

Expert Solution

Explanation of Solution

Given information: f(x)=5cosπx−4

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 a≤c≤b such that f(c)=y .

Calculation:

Consider the equation, f(x)=5cosπx−4

Substituting x=1 in f(x)=5cosπx−4 , we get

f(1)=5cosπ−4=5(−1)−4=−5−4=−9

Substituting x=0 in f(x)=5cosπx−4 , we get

f(0)=5cosπ(0)−4=5−4=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)=8x4−8x2+1 .

c.

Expert Solution

Explanation of Solution

Given information: f(x)=8x4−8x2+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 a≤c≤b such that f(c)=y .

Calculation:

Consider the equation, f(x)=8x4−8x2+1

Substituting x=1 in f(x)=8x4−8x2+1 , we get

f(1)=8(1)4−8(1)2+1=8−8+1=1

Substituting x=0 in f(x)=8x4−8x2+1 , we get

f(0)=8(0)4−8(0)2+1=0−0+1=1

Substituting x = 12 in f(x)=8x4−8x2+1, we get

f(12)=8( 1 2)4−8( 1 2)2+1=8(1 16)−8(14)+1=12−2+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.

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Answer 2

Question

Chapter 0.5, Problem 1E

a.

To determine

To prove: f(c)=0 for some 0<c<1 : f(x)=x3−4x+1 .

a.

Expert Solution

Explanation of Solution

Given information: f(x)=x3−4x+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 a≤c≤b such that f(c)=y .

Calculation:

Consider the equation, f(x)=x3−4x+1

Substituting x=1 in f(x)=x3−4x+1 , we get

f(1)=13−4(1)+1=1−4+1=−2

Substituting x=0 in f(x)=x3−4x+1 , we get

f(0)=03−4(0)+1=0−0+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)=5cosπx−4 .

b.

Expert Solution

Explanation of Solution

Given information: f(x)=5cosπx−4

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 a≤c≤b such that f(c)=y .

Calculation:

Consider the equation, f(x)=5cosπx−4

Substituting x=1 in f(x)=5cosπx−4 , we get

f(1)=5cosπ−4=5(−1)−4=−5−4=−9

Substituting x=0 in f(x)=5cosπx−4 , we get

f(0)=5cosπ(0)−4=5−4=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)=8x4−8x2+1 .

c.

Expert Solution

Explanation of Solution

Given information: f(x)=8x4−8x2+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 a≤c≤b such that f(c)=y .

Calculation:

Consider the equation, f(x)=8x4−8x2+1

Substituting x=1 in f(x)=8x4−8x2+1 , we get

f(1)=8(1)4−8(1)2+1=8−8+1=1

Substituting x=0 in f(x)=8x4−8x2+1 , we get

f(0)=8(0)4−8(0)2+1=0−0+1=1

Substituting x = 12 in f(x)=8x4−8x2+1, we get

f(12)=8( 1 2)4−8( 1 2)2+1=8(1 16)−8(14)+1=12−2+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.

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Answer 3

Question

Chapter 0.5, Problem 1E

a.

To determine

To prove: f(c)=0 for some 0<c<1 : f(x)=x3−4x+1 .

a.

Expert Solution

Explanation of Solution

Given information: f(x)=x3−4x+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 a≤c≤b such that f(c)=y .

Calculation:

Consider the equation, f(x)=x3−4x+1

Substituting x=1 in f(x)=x3−4x+1 , we get

f(1)=13−4(1)+1=1−4+1=−2

Substituting x=0 in f(x)=x3−4x+1 , we get

f(0)=03−4(0)+1=0−0+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)=5cosπx−4 .

b.

Expert Solution

Explanation of Solution

Given information: f(x)=5cosπx−4

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 a≤c≤b such that f(c)=y .

Calculation:

Consider the equation, f(x)=5cosπx−4

Substituting x=1 in f(x)=5cosπx−4 , we get

f(1)=5cosπ−4=5(−1)−4=−5−4=−9

Substituting x=0 in f(x)=5cosπx−4 , we get

f(0)=5cosπ(0)−4=5−4=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)=8x4−8x2+1 .

c.

Expert Solution

Explanation of Solution

Given information: f(x)=8x4−8x2+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 a≤c≤b such that f(c)=y .

Calculation:

Consider the equation, f(x)=8x4−8x2+1

Substituting x=1 in f(x)=8x4−8x2+1 , we get

f(1)=8(1)4−8(1)2+1=8−8+1=1

Substituting x=0 in f(x)=8x4−8x2+1 , we get

f(0)=8(0)4−8(0)2+1=0−0+1=1

Substituting x = 12 in f(x)=8x4−8x2+1, we get

f(12)=8( 1 2)4−8( 1 2)2+1=8(1 16)−8(14)+1=12−2+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.

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Answer 4

Question

Chapter 0.5, Problem 1E

a.

To determine

To prove: f(c)=0 for some 0<c<1 : f(x)=x3−4x+1 .

a.

Expert Solution

Explanation of Solution

Given information: f(x)=x3−4x+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 a≤c≤b such that f(c)=y .

Calculation:

Consider the equation, f(x)=x3−4x+1

Substituting x=1 in f(x)=x3−4x+1 , we get

f(1)=13−4(1)+1=1−4+1=−2

Substituting x=0 in f(x)=x3−4x+1 , we get

f(0)=03−4(0)+1=0−0+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)=5cosπx−4 .

b.

Expert Solution

Explanation of Solution

Given information: f(x)=5cosπx−4

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 a≤c≤b such that f(c)=y .

Calculation:

Consider the equation, f(x)=5cosπx−4

Substituting x=1 in f(x)=5cosπx−4 , we get

f(1)=5cosπ−4=5(−1)−4=−5−4=−9

Substituting x=0 in f(x)=5cosπx−4 , we get

f(0)=5cosπ(0)−4=5−4=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)=8x4−8x2+1 .

c.

Expert Solution

Explanation of Solution

Given information: f(x)=8x4−8x2+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 a≤c≤b such that f(c)=y .

Calculation:

Consider the equation, f(x)=8x4−8x2+1

Substituting x=1 in f(x)=8x4−8x2+1 , we get

f(1)=8(1)4−8(1)2+1=8−8+1=1

Substituting x=0 in f(x)=8x4−8x2+1 , we get

f(0)=8(0)4−8(0)2+1=0−0+1=1

Substituting x = 12 in f(x)=8x4−8x2+1, we get

f(12)=8( 1 2)4−8( 1 2)2+1=8(1 16)−8(14)+1=12−2+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.

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Answer 5

Chapter 0.5, Problem 1E

a.

To determine

To prove: f(c)=0 for some 0<c<1 : f(x)=x3−4x+1 .

a.

Expert Solution

Explanation of Solution

Given information: f(x)=x3−4x+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 a≤c≤b such that f(c)=y .

Calculation:

Consider the equation, f(x)=x3−4x+1

Substituting x=1 in f(x)=x3−4x+1 , we get

f(1)=13−4(1)+1=1−4+1=−2

Substituting x=0 in f(x)=x3−4x+1 , we get

f(0)=03−4(0)+1=0−0+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)=5cosπx−4 .

b.

Expert Solution

Explanation of Solution

Given information: f(x)=5cosπx−4

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 a≤c≤b such that f(c)=y .

Calculation:

Consider the equation, f(x)=5cosπx−4

Substituting x=1 in f(x)=5cosπx−4 , we get

f(1)=5cosπ−4=5(−1)−4=−5−4=−9

Substituting x=0 in f(x)=5cosπx−4 , we get

f(0)=5cosπ(0)−4=5−4=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)=8x4−8x2+1 .

c.

Expert Solution

Explanation of Solution

Given information: f(x)=8x4−8x2+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 a≤c≤b such that f(c)=y .

Calculation:

Consider the equation, f(x)=8x4−8x2+1

Substituting x=1 in f(x)=8x4−8x2+1 , we get

f(1)=8(1)4−8(1)2+1=8−8+1=1

Substituting x=0 in f(x)=8x4−8x2+1 , we get

f(0)=8(0)4−8(0)2+1=0−0+1=1

Substituting x = 12 in f(x)=8x4−8x2+1, we get

f(12)=8( 1 2)4−8( 1 2)2+1=8(1 16)−8(14)+1=12−2+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.

Answer 6

Chapter 0.5, Problem 1E

a.

To determine

To prove: f(c)=0 for some 0<c<1 : f(x)=x3−4x+1 .

a.

Expert Solution

Explanation of Solution

Given information: f(x)=x3−4x+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 a≤c≤b such that f(c)=y .

Calculation:

Consider the equation, f(x)=x3−4x+1

Substituting x=1 in f(x)=x3−4x+1 , we get

f(1)=13−4(1)+1=1−4+1=−2

Substituting x=0 in f(x)=x3−4x+1 , we get

f(0)=03−4(0)+1=0−0+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.

Answer 7

Chapter 0.5, Problem 1E

a.

To determine

To prove: f(c)=0 for some 0<c<1 : f(x)=x3−4x+1 .

Answer 8

a.

Expert Solution

Explanation of Solution

Given information: f(x)=x3−4x+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 a≤c≤b such that f(c)=y .

Calculation:

Consider the equation, f(x)=x3−4x+1

Substituting x=1 in f(x)=x3−4x+1 , we get

f(1)=13−4(1)+1=1−4+1=−2

Substituting x=0 in f(x)=x3−4x+1 , we get

f(0)=03−4(0)+1=0−0+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.

Answer 9

a.

Expert Solution

Answer 10

a.

Expert Solution

Answer 11

Expert Solution

Answer 12

b.

To determine

To prove: f(c)=0 for some 0<c<1 : f(x)=5cosπx−4 .

b.

Expert Solution

Explanation of Solution

Given information: f(x)=5cosπx−4

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 a≤c≤b such that f(c)=y .

Calculation:

Consider the equation, f(x)=5cosπx−4

Substituting x=1 in f(x)=5cosπx−4 , we get

f(1)=5cosπ−4=5(−1)−4=−5−4=−9

Substituting x=0 in f(x)=5cosπx−4 , we get

f(0)=5cosπ(0)−4=5−4=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.

Answer 13

b.

To determine

To prove: f(c)=0 for some 0<c<1 : f(x)=5cosπx−4 .

Answer 14

b.

Expert Solution

Explanation of Solution

Given information: f(x)=5cosπx−4

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 a≤c≤b such that f(c)=y .

Calculation:

Consider the equation, f(x)=5cosπx−4

Substituting x=1 in f(x)=5cosπx−4 , we get

f(1)=5cosπ−4=5(−1)−4=−5−4=−9

Substituting x=0 in f(x)=5cosπx−4 , we get

f(0)=5cosπ(0)−4=5−4=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.

Answer 15

b.

Expert Solution

Answer 16

b.

Expert Solution

Answer 17

Expert Solution

Answer 18

c.

To determine

To prove: f(c)=0 for some 0<c<1 : f(x)=8x4−8x2+1 .

c.

Expert Solution

Explanation of Solution

Given information: f(x)=8x4−8x2+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 a≤c≤b such that f(c)=y .

Calculation:

Consider the equation, f(x)=8x4−8x2+1

Substituting x=1 in f(x)=8x4−8x2+1 , we get

f(1)=8(1)4−8(1)2+1=8−8+1=1

Substituting x=0 in f(x)=8x4−8x2+1 , we get

f(0)=8(0)4−8(0)2+1=0−0+1=1

Substituting x = 12 in f(x)=8x4−8x2+1, we get

f(12)=8( 1 2)4−8( 1 2)2+1=8(1 16)−8(14)+1=12−2+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.

Answer 19

c.

To determine

To prove: f(c)=0 for some 0<c<1 : f(x)=8x4−8x2+1 .

Answer 20

c.

Expert Solution

Explanation of Solution

Given information: f(x)=8x4−8x2+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 a≤c≤b such that f(c)=y .