*Remarks in italics are by me for extra explanation. These comments would not be necessary for full marks in an exam situation. Exercises taken from p.75 of these notes.*

## Question 1

*If , show that there is a number such that .*

### Solution

Solving the equation is equivalent to finding a root of the continuous function . Note that this function is continuous hence we can use the Intermediate Value Theorem to find an interval with a root.

,

,

,

.

Hence changes sign between 2 and 3. Hence has a root in , say at . Hence

## Question 2

*Use the Intermediate Value Theorem to prove that there is positive number such that (this proves existence of the number ).*

### Solution

Let . Now

, , .

Hence has a root . That is

## Question 3

*Use the Intermediate Value Theorem to show that there is a root of the given equation in the specified interval (i) , (ii) , (iii) , (iv) , *

### Solution

#### Part (i)

Let . Now and . Hence has a root in .

#### Part (ii)

Let . Now and . Hence has a root in .

#### Part (iii)

Let . Now , . Hence has a root in .

#### Part (iv)

Let . Now , so in fact is a root (**obviously I’ve made a typo here!**). However we want to look for a root in — this interval does not included . Hence look at the interval . Now , . Hence has a root in — and thus in .

## Question 4

*Use the Intermediate Value Theorem to locate an interval of length in which each of the following equations have a roots (note that in general a polynomial of degree has roots — I just want ye to find a location of one of them.). *

*.**.**.**.**.**.**.**.*

*Now use the Bisection Method to find intervals of length less than (this will require four iterations of the Bisection Method — after four iterations the interval on which we know there is a root will have length ).*

*Now use the Newton Method to find a root of (i) to 1 decimal place, (ii) to two decimal places (iii) to three decimals… (viii) to 8 decimal places.*

### Solution

#### Part (i)

Let . Now , , . Hence there is a root in .

- Now using the bisection method evaluate . Hence there is a root in (because is negative at 1 and positive at 1.5).
- Hence evaluate . Hence there is a root in .
- Hence evaluate . Hence there is a root in .
- Hence evaluate . Hence there is a root in .

Now (*we may as well*) we can use the midpoint of this interval as our seed for the Newton-Raphson Method, . Once we get consecutive agreement to one decimal place we will finish.

Now . Hence,

,

.

#### Part (ii)

Let . Note that for all , . Hence we have to look for a root in the negative real numbers. Now , . Hence there is a root in .

- By the Bisection Method, we evaluate . Hence there is a root in .
- Evaluate . Hence there is a root in .
- Evaluate . Hence there is a root in .
- Evaluate . Hence there is a root in .

Letting .

.

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