Numerical Methods for Engineers
Numerical Methods for Engineers
7th Edition
ISBN: 9780073397924
Author: Steven C. Chapra Dr., Raymond P. Canale
Publisher: McGraw-Hill Education
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Chapter 6, Problem 2P

Determine the highest real root of

f ( x ) = 2 x 3 − 11.7 x 2 + 17.7 x − 5

(a) Graphically.

(b) Fixed-point iteration method (three iterations, x 0 = 3 ). Note: Make certain that you develop a solution that converges on the root.

(c) Newton-Raphson method (three iterations, x 0 = 3 ).

(d) Secant method (three iterations, x − 1 = 3 ,     x 0 = 4 ).

(e) Modified secant method (three iterations, x 0 = 3 ,     δ = 0.01 ).

Compute the approximate percent relative errors for your solutions.

(a)

Expert Solution
Check Mark
To determine

To calculate: The highest real root of the function f(x)=2x3−11.7x2+17.7x−5 by the use of graphical method.

Answer to Problem 2P

Solution:

The highest real root of the function f(x)=2x3−11.7x2+17.7x−5 is 3.6.

Explanation of Solution

Given:

The function, f(x)=2x3−11.7x2+17.7x−5.

Formula used:

The roots of the function are the points at which the graph of the function crosses the x-axis.

Calculation:

Consider the function,

f(x)=2x3−11.7x2+17.7x−5

Substitute different values of x and find the corresponding values of f(x) to graph the function f(x)=2x3−11.7x2+17.7x−5,

For x=0,

f(0)=2(0)3−11.7(0)2+17.7(0)−5=0−0+0−5=−5

For x=1,

f(1)=2(1)3−11.7(1)2+17.7(1)−5=2−11.7+17.7−5=19.7−16.7=3

For x=2,

f(2)=2(2)3−11.7(2)2+17.7(2)−5=16−46.8+35.4−5=51.4−51.8=−0.4

For x=3,

f(3)=2(3)3−11.7(3)2+17.7(3)−5=54−105.3+53.1−5=107.1−110.3=−3.2

For x=4,

f(4)=2(4)3−11.7(4)2+17.7(4)−5=128−187.2+70.8−5=198.8−192.2=6.6

Summarize the above values as shown below,

x f(x)=2x3−11.7x2+17.7x−5
0 −5
1 3
2 −0.4
3 −3.2
4 6.6

Plot the above points on the graph and join them as below,

Numerical Methods for Engineers, Chapter 6, Problem 2P

From the above graph, it is observed that the graph of the function crosses the x-axis from three points. That is, approximately x=0.35, x=1.9 and x=3.6.

Hence, the highest real root of the function f(x)=2x3−11.7x2+17.7x−5 is 3.6.

(b)

Expert Solution
Check Mark
To determine

To calculate: The highest real root of the function f(x)=2x3−11.7x2+17.7x−5 by the use of fixed point iteration method with the initial guess x0=3 and up to three iterations.

Answer to Problem 2P

Solution:

The highest real root of the function f(x)=2x3−11.7x2+17.7x−5 is 3.4425.

Explanation of Solution

Given:

The function, f(x)=2x3−11.7x2+17.7x−5.

Formula used:

The simple fixed-point iteration formula for the function x=g(x),

xi+1=g(xi)

And, formula for approximate error is,

εa=|xi+1−xixi+1|100%

Calculation:

Consider the function,

f(x)=2x3−11.7x2+17.7x−5

The function can be formulated as fixed-point iteration as,

2x3−11.7x2+17.7x−5=017.7x=−2x3+11.7x2+5xi+1=−2xi3+11.7xi2+517.7

Use initial guess x0=3, the first iteration is,

x0+1=−2x03+11.7x02+517.7x1=−2(3)3+11.7(3)2+517.7=−54+105.3+517.7=3.18079

Therefore, the approximate error is,

εa=|3.18079−33.18079|×100%=|0.180793.18079|×100%=|0.05683|×100%=5.683%

Use x1=3.18079, the second iteration is,

x1+1=−2x13+11.7x12+517.7x2=−2(3.18079)3+11.7(3.18079)2+517.7=−64.3628+118.3739+517.7=3.333959

Therefore, the approximate error is,

εa=|3.333959−3.180793.333959|×100%=|0.1531693.333959|×100%=|0.04594|×100%=4.594%

Use x2=3.333959, the third iteration is,

x2+1=−2x23+11.7x22+517.7x3=−2(3.333959)3+11.7(3.333959)2+517.7=−74.1158+130.0488+517.7=3.4425

Therefore, the approximate error is,

εa=|3.4425−3.3339593.4425|×100%=|0.1085413.4425|×100%=|0.0315297|×100%=3.153%

Thus, all the iteration can be summarized as below,

i xi εa=|xi+1−xixi+1|100%
0 3
1 3.18079 5.683%
2 3.333959 4.594%
3 3.4425 3.153%

Hence, the highest root is 3.4425.

(c)

Expert Solution
Check Mark
To determine

To calculate: The highest real root of the function f(x)=2x3−11.7x2+17.7x−5 by the use of Newton-Raphson method with the initial guess x0=3 and up to three iterations.

Answer to Problem 2P

Solution:

The highest real root of the function f(x)=2x3−11.7x2+17.7x−5 is 3.792837.

Explanation of Solution

Given:

The function, f(x)=2x3−11.7x2+17.7x−5.

Formula used:

The Newton-Raphson formula,

xi+1=xi−f(xi)f′(xi)

And, formula for approximate error is,

εa=|xi+1−xixi+1|100%

Calculation:

Consider the function,

f(x)=2x3−11.7x2+17.7x−5

Differentiate the above function with respect to x,

f′(x)=ddx(2x3−11.7x2+17.7x−5)=ddx(2x3)+ddx(−11.7x2)+ddx(17.7x)+ddx(−5)=2×3x2−11.7×2x+17.7−0=6x2−23.4x+17.7

The initial guess is x0=3, thus the first iteration is,

x0+1=x0−f(x0)f′(x0)x1=3−{2(3)3−11.7(3)2+17.7(3)−5}{6(3)2−23.4(3)+17.7}=3−(−3.21.5)=5.133

Therefore, the approximate error is,

εa=|5.133−35.133|×100%=|2.1335.133|×100%=|0.41555|×100%=41.555%

Use x1=5.133, the second iteration is,

x1+1=x1−f(x1)f′(x1)x2=5.133−{2(5.133)3−11.7(5.133)2+17.7(5.133)−5}{6(5.133)2−23.4(5.133)+17.7}=5.133−(48.071555.6739)=4.26955

Therefore, the approximate error is,

εa=|4.26955−5.1334.26955|×100%=|−0.863454.26955|×100%=|−0.20223|×100%=20.223%

Use x2=4.26955, the third iteration is,

x2+1=x2−f(x2)f′(x2)x3=4.26955−{2(4.26955)3−11.7(4.26955)2+17.7(4.26955)−5}{6(4.26955)2−23.4(4.26955)+17.7}=4.26955−(12.950827.16687)=3.792837

Therefore, the approximate error is,

εa=|3.792837−4.269553.792837|×100%=|−0.4767133.792837|×100%=|−0.12569|×100%=12.569%

Similarly, all the iteration can be summarized as below,

i xi εa=|xi+1−xixi+1|100%
0 3
1 5.133 41.555%
2 4.26955 20.223%
3 3.792837 12.569%

Hence, the highest root is 3.792837.

(d)

Expert Solution
Check Mark
To determine

To calculate: The highest real root of the function f(x)=2x3−11.7x2+17.7x−5 by the use of Secant method with the initial guess x−1=3 and x0=4 and up to three iterations.

Answer to Problem 2P

Solution:

The highest real root of the function f(x)=2x3−11.7x2+17.7x−5 is 3.58629.

Explanation of Solution

Given:

The function, f(x)=2x3−11.7x2+17.7x−5.

Formula used:

The iterative equation of secant method is,

xi+1=xi−f(xi)(xi−1−xi)f(xi−1)−f(xi)

And, formula for approximate error is,

εa=|xi+1−xixi+1|100%

Calculation:

Consider the function,

f(x)=2x3−11.7x2+17.7x−5

The initial guess is x−1=3 and x0=4, thus the first iteration is,

x0+1=x0−f(x0)(x0−1−x0)f(x0−1)−f(x0)x1=4−[{2(4)3−11.7(4)2+17.7(4)−5}(3−4)][{2(3)3−11.7(3)2+17.7(3)−5}−{2(4)3−11.7(4)2+17.7(4)−5}]=4−[{6.6}(−1)][{−3.2}−{6.6}]=3.3265

Therefore, the approximate error is,

εa=|3.3265−43.3265|×100%=|−0.67353.3265|×100%=|−0.2025|×100%=20.25%

Use x1=3.3265, the second iteration is,

x1+1=x1−f(x1)(x1−1−x1)f(x1−1)−f(x1)x2=3.3265−[{2(3.3265)3−11.7(3.3265)2+17.7(3.3265)−5}(4−3.3265)][{2(4)3−11.7(4)2+17.7(4)−5}−{2(3.3265)3−11.7(3.3265)2+17.7(3.3265)−5}]=3.3265−[{−1.969}(0.6735)][{6.6}−{−1.969}]=3.4812

Therefore, the approximate error is,

εa=|3.4812−3.32653.4812|×100%=|0.15743.4812|×100%=|0.04443|×100%=4.443%

Use x2=3.4812, the third iteration is,

x2+1=x2−f(x2)(x2−1−x2)f(x2−1)−f(x2)x3=3.4812−[{2(3.4812)3−11.7(3.4812)2+17.7(3.4812)−5}(3.3265−3.4812)][{2(3.3265)3−11.7(3.3265)2+17.7(3.3265)−5}−{2(3.4812)3−11.7(3.4812)2+17.7(3.4812)−5}]=3.4812−[{−0.7965}(−0.1547)][{−1.969}−{−0.7965}]=3.58629

Therefore, the approximate error is,

εa=|3.58629−3.48123.58629|×100%=|0.105093.58629|×100%=|0.02930|×100%=2.93%

Similarly, all the iteration can be summarized as below,

i xi εa=|xi+1−xixi+1|100%
0 4
1 3.3265 20.25%
2 3.4812 4.443%
3 3.58629 2.93%

Hence, the highest root is 3.58629.

(e)

Expert Solution
Check Mark
To determine

To calculate: The highest real root of the function f(x)=2x3−11.7x2+17.7x−5 by the use of modified Secant method with the initial guess x0=3 and δ=0.01 and up to three iterations.

Answer to Problem 2P

Solution:

The highest real root of the function f(x)=2x3−11.7x2+17.7x−5 is 3.7429.

Explanation of Solution

Given:

The function, f(x)=2x3−11.7x2+17.7x−5.

Formula used:

The iteration formula for modified secant method is,

xi+1=xi−δxif(xi)f(xi+δxi)−f(xi)

And, formula for approximate error is,

εa=|xi+1−xixi+1|100%

Calculation:

Consider the function,

f(x)=2x3−11.7x2+17.7x−5

Use initial guess of x0=3 and δ=0.01, the first iteration is,

x0+1=x0−δx0f(x0)f(x0+δx0)−f(x0)x1=3−0.01×3×f(3)f(3+0.01×3)−f(3)=3−0.03{2(3)3−11.7(3)2+17.7(3)−5}{2(3.03)3−11.7(3.03)2+17.7(3.03)−5}−{2(3)3−11.7(3)2+17.7(3)−5}=3−0.03×{−3.2}{−3.149276}−{−3.2}

Simplify furthermore,

x1=3−(−0.096)(0.050724)=3−(−1.89259)=3+1.89259=4.89259

Therefore, the approximate error is,

εa=|4.89259−34.89259|×100%=|1.892594.89259|×100%=38.68%

Use x1=4.89259 and δ=0.01, the second iteration is,

x1+1=x1−δx1f(x1)f(x1+δx1)−f(x1)x2=4.89259−0.01×4.89259×f(4.89259)f(4.89259+0.01×4.89259)−f(4.89259)=4.89259−0.049{2(4.89259)3−11.7(4.89259)2+17.7(4.89259)−5}{2(4.9415)3−11.7(4.9415)2+17.7(4.9415)−5}−{2(4.89259)3−11.7(4.89259)2+17.7(4.89259)−5}=4.89259−0.049×{35.763}{38.096}−{35.763}

Simplify furthermore,

x2=4.89259−0.049×{35.763}{38.096}−{35.763}=4.89259−1.75242.333=4.89259−0.75114=4.14145

Therefore, the approximate error is,

εa=|4.14145−4.892594.14145|×100%=|−0.751144.14145|×100%=18.14%

Use x2=4.14145 and δ=0.01, the third iteration is,

x2+1=x2−δx2f(x2)f(x2+δx2)−f(x2)x3=4.14145−0.01×4.14145×f(4.14145)f(4.14145+0.01×4.14145)−f(4.14145)=4.14145−0.041415{2(4.14145)3−11.7(4.14145)2+17.7(4.14145)−5}{2(4.183)3−11.7(4.183)2+17.7(4.183)−5}−{2(4.14145)3−11.7(4.14145)2+17.7(4.14145)−5}=4.14145−0.041415×{9.6949}{10.7025}−{9.6949}

Simplify furthermore,

x3=4.14145−0.041415×{9.6949}{10.7025}−{9.6949}=4.14145−0.40151.0076=4.14145−0.3985=3.7429

Therefore, the approximate error is,

εa=|3.7429−4.141453.7429|×100%=|−0.398553.7429|×100%=10.65%

Similarly, all the iteration can be summarized as below,

i xi εa=|xi+1−xixi+1|100%
0 3
1 4.89259 38.68%
2 4.14145 18.14%
3 3.7429 10.65%

Hence, the highest root is 3.7429.

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