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 $f(x)=x^3-x^2+x$, show that there is a number $c$ such that $f(c)=10$.

### Solution

Solving the equation $f(x)=10$ is equivalent to finding a root of the continuous function $g(x)=f(x)-10=x^3-x^2+x-10$. Note that this function is continuous hence we can use the Intermediate Value Theorem to find an interval with a root.

$g(0)=-10<0$,

$g(1)=(1)^3-(1)^2+(1)-10=-9<0$,

$g(2)=(2)^3-(2)^2+(2)-10=-4<0$,

$g(3)=(3)^3-(3)^2+(3)-10=11>0$.

Hence $g$ changes sign between 2 and 3. Hence $g$ has a root in $(2,3)$, say at $c$. Hence $g(c)=0\Leftrightarrow f(c)=10$ $\bullet$

## Question 2

Use the Intermediate Value Theorem to prove that there is positive number $c$ such that $c^2=2$ (this proves existence of the number $\sqrt{2}$).

### Solution

Let $f(x)=x^2-2$. Now

$f(0)=-2$$f(1)=-1$$f(2)=+2$.

Hence $f$ has a root $c\in (1,2)$. That is $f(c)=0 \Leftrightarrow c^2=2$ $\bullet$

## Question 3

Use the Intermediate Value Theorem to show that there is a root of the given equation in the specified interval (i) $x^4+x-3=0$, $(1,2)$ (ii) $\sqrt[3]{x}=1-x$, $(0,2)$ (iii) $\cos x=x$, $(0,1)$ (iv) $\tan x=2x$, $(0,1.4)$

### Solution

#### Part (i)

Let $f(x)=x^4+x-3$. Now $f(1)=-1$ and $f(2)=15$. Hence $f$ has a root in $(1,2)$.

#### Part (ii)

Let $f(x)=\sqrt{3}x-1+x$. Now $f(0)=-1$ and $f(2)=2\sqrt{3}+1>0$. Hence $f$ has a root in $(0,2)$.

#### Part (iii)

Let $f(x)=\cos x-x$. Now $f(0)=1$$f(1)=\cos 1-1<0$. Hence $f$ has a root in $(0,1)$.

#### Part (iv)

Let $f(x)=\tan x-2x$. Now $f(0)=0$, so in fact $x=0$ is a root (obviously I’ve made a typo here!). However we want to look for a root in $(0,1.4)$ — this interval does not included $0$. Hence look at the interval $(1,1.4)$. Now $f(1)=\tan 1-2\approx-0.443<0$$f(1.4)=\tan(1.4)-2.8\approx 3>0$. Hence $f$ has a root in $(1,1.4)$ — and thus in $(0,1.4)$.

## Question 4

Use the Intermediate Value Theorem to locate an interval of length $1$ in which each of the following equations have a roots (note that in general a polynomial of degree $n$ has $n$ roots — I just want ye to find a location of one of them.).

1. $x^3+2x-4=0$.
2. $x^5+2=0$.
3. $x^3=30$.
4. $x^4+x-4=0$.
5. $x^4=1+x$.
6. $\sqrt{x+3}=x^2$.
7. $x^5-x^4-5x^3-x^2+4x+3=0$.
8. $3\sin (x^2)=2x$.

Now use the Bisection Method to find intervals of length less than $0.1$ (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 $1/2^4=1/16<0.1$).

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 $f(x)=x^3+2x-4$. Now $f(0)=-4<0$$f(1)=-1<0$$f(2)=8>0$. Hence there is a root in $(1,2)$.

1. Now using the bisection method evaluate $f(1.5)=2.375>0$. Hence there is a root in $(1,1.5)$ (because $f$ is negative at 1 and positive at 1.5).
2. Hence evaluate $f(1.25)\approx 0.453>0$. Hence there is a root in $(1,1.25)$.
3. Hence evaluate $f(1.125)\approx -0.326$. Hence there is a root in $(1.125,1.25)$.
4. Hence evaluate $f(1.1875)\approx 0.05>0$. Hence there is a root in $(1.125,1.1875)$.

Now (we may as well) we can use the midpoint of this interval as our seed for the Newton-Raphson Method, $x_0=1.15625$. Once we get consecutive agreement to one decimal place we will finish.

Now $f'(x)=3x^2+2$. Hence,

$x_1=x_0-\frac{f(x_0)}{f'(x_0)}$,

$\Rightarrow x_1=0.15625-\frac{-0.141693}{6.01074}=1.17982$.

#### Part (ii)

Let $f(x)=x^5+2$. Note that for all $x>0$$f(x)>0$. Hence we have to look for a root in the negative real numbers. Now $f(-1)=1>0$$f(-2)=-30<0$. Hence there is a root in $(-2,-1)$.

1. By the Bisection Method, we evaluate $f(-1.5)\approx -5.6<0$. Hence there is a root in $(-1.5,-1)$.
2. Evaluate $f(-1.25)\approx -1.05$. Hence there is a root in $(-1.25,-1)$.
3. Evaluate $f(-1.125)\approx 0.2>0$. Hence there is a root in $(-1.25,-1.125)$.
4. Evaluate $f(-1.1875)\approx -0.36$. Hence there is a root in $(-1.1875,-1.125)$.

Letting $x_0=-1.15625$.

$x_1=x_0-\frac{f(x_0)}{f'(x_0)}=-1.149$

$x_2=x_1-\frac{f(x_1)}{f'(x_1)}=-1.149$.