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)=-2f(1)=-1f(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)=1f(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<0f(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<0f(1)=-1<0f(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>0f(x)>0. Hence we have to look for a root in the negative real numbers. Now f(-1)=1>0f(-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.

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