Numerical Solutions
Methods for locating roots to a polynomial equation:
a. Locate roots of f(x) =0 by considering changes of sign.
b. Using iteration to find an approximation to the roots of the equation f(x) = 0.
c. Newton_Raphson procedure to find approximation to the root of the equation f(x)=0.
To show solution lies in an interval:
1. The diagram shows a sketch of the curve y = f(x), where f(x) = x3 -4x2 + 3x +1.
a. Explain how the graph shows that f(x) has a root between x=2 and x=3.
b. Show that f(x) has a root between x=1.4 and x= 1.5.
2a.Using the same axes , sketch the graphs of y=lnx and y=1/x. Explain how your diagram shows that the
function f(x) = lnx -1/x has only one roots.
2b. Show that this root lies in the interval 1.7 < x < 1.8,
2c. Given that the root of f(x) is α , show that α=1.763 correct to 3 dp.
3.Show that each of these functions has at least one root in the given interval.
a. f(x) = x3 –x + 5 , -2<x<-1
x
b. f(x) = e =lnx – 5 , 1.65 < x < 1.75
4. h(x) = 3√x – cosx – 1 , where x is in radians.
a. Show that the equation h(x) has a roots , α , between x = 1.4 and x=1.5.
b. By choosing a suitable interval, show that α=1.441 is correct to 3 decimal places.
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5. f(x) = (105x3 -128x2 +49x – 6)cos2x, where x is in radians . The diagram below shows the sketch of
y= f(x).
a. Calculate f(0.2) and f(0.8).
b. Use your answer to part a to make a conclusion about the number of roots of f(x) in the interval 0.2 < x
< 0.8 .
c. Further calculate f(0.3), f(0.4, f(0.5) and f(0.7).
d. Use your answer to parts a and c to make an improved conclusion about the number of roots of f(x) in
the interval 0.2 < x < 0.8.
6. h(x) = sin2x + e4x
a. Show that there is a stationary point , α , of y=h(x) in the interval -0.9 < x < -0.8.
b. By considering the change of sign of h’(x) in a suitable interval, verify that α =-0.823 correct to 3 dp.
Iteration:
To solve an equation of the form f(x) = 0 by an iterative method, rearrange f(x) =0 into form x=g(x) and
use the iterative formula xn+1 = g(xn).
Graphically these iterations lead to staircase diagrams.
8. f(x) = x2 – 4x + 1.
a. Show that the equation f(x) = 0 can be written as x = 4 – 1/x , x≠ 0.
f(x) have a root α, in the interval 3< α < 4.
b. Use the iterative formula xn+1 = 4 – 1/xn with x0 = 3 to find the value of x1, x2, x3.
9. f(x) = x3 – 3x2 – 2x + 5
a. Show that the eqn f(x) = 0 has a root in the interval 3 < x < 4.
b. Use the iterative formula xn+1 = √ xn3 – 2xn + 5 { Note square root is for the whole expression}
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to calculate the values of x1, x2, x3 , giving your answer to 4 dp and taking :
i. x0 = 1.5 ii. x0 = 4
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10. f(x) = x2 – 6x + 1
a. Show that the eqn f(x) = 0 can be written as x= √ (6x-1).
b. Sketch on the same axes the graphs of y=x and y = √(6x-1).
c. Write down the number of roots of f(x).
d. Use your diagram to explain why the iterative formula xn+1 = √(6x – 1) converges to a root of f(x)
when x0 = 2.
f(x) = 0 can also be rearranged to form the iterative formula xn+1 = xn2 + 1
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e. By sketching a diagram explain why the iteration diverges when x0 = 10.
11. f(x) = xe-x –x + 2
a. Show that the equation f(x) = 0 can be written as x = ln | x | , x ≠ 2
| x-2 |
f(x) has a root , α , in the interval -2 < x < -1.
b. Use iterative formula xn+1 = ln | xn | , x ≠ 2 with x0 = -1 to find , to 2 decimal
| x-2 |
places, the value of x1 , x2 and x3.
12. f(x) = 3cos(x2) + x – 2
a. Show that the equation f(x) = 0 can be written as x = (arccos 2-x )1/2
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b. Use the iterative formula xn+1 = (arccos 2 – xn )1/2 , x0 = 1 to find , to 3 decimal
3
places the value of x1 , x2 and x3.
c. Given that f(x) = 0 has only one root, α , show that α = 1.1298 correct to 4 dp.
13. The diagram shows a sketch of part of the curve with equation y=f(x), where
f(x) = xex – 4x. The curve cuts the x-axis at the points A and B and has a minimum points at P, as shown
in the diagram.
A
O B
P
a. Workout the co ordinates of A and the coordinates of B.
b. Find f’(x).
c. Show that the x coordinate of P lies between 0.7 and 0.8.
d. Show that the x coordinate of P is the solution to the equation x = ln 4 .
x+1
To find an approximation for the x co-ordinate of P, iterative formula
xn+1 = ln 4 is used.
xn + 1
e. Let x0 = 0. Find the values of x1 , x2 and x3 to 3 decimal places.
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Newton-Raphson formula
xn+1 = xn - f(x n )
f ‘(xn )
14. The diagram shows part of the curve with equation y=f(x), where
f(x) = x3 +2x2-5x – 4. The point A , with x coordinate p , is a stationary point on the curve.
The equation f(x) = 0 has a root , α , in the interval 1.8 < x < 1.9.
a. Explain why x0 = p is not suitable to use as a first approximation to α when applying Newton-Raphson
method to f(x).
b. Using x0=2 as a first approximation to α , apply the Newton-Raphson procedure twice to f(x) to find a
second approximation to α, giving your answer to 3 decimal places.
c. By considering the change of sign in f(x) over an appropriate interval, show that your answer to part b
is accurate to 3 decimal place.
15. f(x) = x2 – 4/x – 10, x≠ 0.
a. Use differentiation to find f ’(x) .
The roots α , of the eqn f(x) = 0 lies in the interval [-0.4, -0.3].
b. Taking -0.4 as a first approximation to α , apply Newton-Raphson process once to f(x) to obtain a
second approximation to α. Give your answer to 3 dp.
16. y = f(x) , where f(x) = x2sinx – 2x +1. The points P, Q, R are roots of the equation.
The points A and B are stationary points , with x-coordinates a and b respectively.
a. Show that the curve has a root in each of the following intervals:
i. [0.6, 0.7] ii. [1.2,1.3] iii. [2.4, 2.5]
b. Explain why x0 = a is not suitable to sue as a first approximation to α when applying the Newton-
Raphson method to f(x).
c. Using x0 =2.4 as a first approximation, apply the Newton-Raphson method to f(x) to obtain a second
approximation. Give your answer to 3 dp.
17a. By writing y = xx in the form lny = xlnx , show that dy/dx = xx(lnx + 1).
b. Show that the function f(x) = xx -2 has a root, α , in the interval [1.4, 1.6].
c. Taking x0 = 1.5 as a first approximation to α, apply Newton-Raphson procedure once to obtain a
second approximation to α, giving your answer to 4 dp.
d. By considering a change of sign of f(x) over a suitable interval, show that
α = 1.5596 , correct to 4 dp.
