Determine the real root of -12- 21x + 18x² - 2.75x³ graphically (choose your limits). In addition, determine the first root of the function with
Determine the real root of -12- 21x + 18x² - 2.75x³ graphically (choose your limits). In addition, determine the first root of the function with
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Question
Can someone graph this for me (with the guide question and answer by an expert)? By drawing. Thank you.
Transcribed Image Text:Determine the real root of -12- 21x + 18x² - 2.75x³
graphically (choose your limits). In addition, determine
the first root of the function with
1) bisection
2) false position
For (1) and (2) use initial guesses of x₁ =
and a stopping criterion of 1%
-1 and xu
Xu = 0,
![Step 1
Given that, f(x) = -12-21x + 18x² - 275x²
+(-1)=297570 and f(0) = -12 <0
So, the root of f(x)=0 lies between [-1,03
So, Xo =
30,
n
-1+0
2
now the root
Step 2
continuing this process of bisection method,
we get the following table-
n
xn= a +b
O
1
-0.5
2 -0.5
3 -0.5
4 -0.9375 08631 -0.375-14487
5 -0.9375 0.8631 -0.4062-0.3137
-0.9210 0·2625-014062-0.3137
-0.42 10 0·2695 -0.9191 -0.0234
8 1-0.918 | 0·12276-0'4141 -0.0234
1-0.416 0·0996 -0.4141 -0.0234
6
7
9
so, approximate root is -0.41 (corrected Upto)
O
1
2
3
4
20:5 & + (-0·5) = 3·343870
between [~0.5, 0]
5
lies
xn
f(a)
f(b)
O
-12
29.75
3.3438
O
-12
3.3438 -0.25 -5.582
3.3438-0375-14487"
Now using faise position method, we will use
the recursive relation, Xn+₂ = xn-
Xn+1-Xn
f(xn)
F(x₂+₁)-f(x₂)
+(Xn+2).
xn+2
-0.5
-0.25
+0.375
-0.4375
-0.4062
-0.4219
-64141
-0.418
-0.416
-6:415
Bisection
f(xn)
33438
-5582
-14487
0.8631
-0.3137
0.2695
-0.0234
61227
09496
0.0131
f(x) x₂+1
f(x+1)
G
29.75
-12 -0.2874-44117
29.75 -0.2874-44117-0.3794 -1:2897
29.75 -6.3794-1·2897-04052-0 35 13
29.75 -04052-0.3513-04122 -0.0938
29:75 -04122-00938 -0.419
-0.0249
29.75 -0414-0·02491-0·41451-0.0066
50, approximate root.
is -0.41.
And false position method converges
faster than
method.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F25561a02-472f-4221-8559-8bee97ef406c%2F56d43cc4-d7a6-4680-949d-8bf7b9e46c66%2Fw3qek92_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Step 1
Given that, f(x) = -12-21x + 18x² - 275x²
+(-1)=297570 and f(0) = -12 <0
So, the root of f(x)=0 lies between [-1,03
So, Xo =
30,
n
-1+0
2
now the root
Step 2
continuing this process of bisection method,
we get the following table-
n
xn= a +b
O
1
-0.5
2 -0.5
3 -0.5
4 -0.9375 08631 -0.375-14487
5 -0.9375 0.8631 -0.4062-0.3137
-0.9210 0·2625-014062-0.3137
-0.42 10 0·2695 -0.9191 -0.0234
8 1-0.918 | 0·12276-0'4141 -0.0234
1-0.416 0·0996 -0.4141 -0.0234
6
7
9
so, approximate root is -0.41 (corrected Upto)
O
1
2
3
4
20:5 & + (-0·5) = 3·343870
between [~0.5, 0]
5
lies
xn
f(a)
f(b)
O
-12
29.75
3.3438
O
-12
3.3438 -0.25 -5.582
3.3438-0375-14487"
Now using faise position method, we will use
the recursive relation, Xn+₂ = xn-
Xn+1-Xn
f(xn)
F(x₂+₁)-f(x₂)
+(Xn+2).
xn+2
-0.5
-0.25
+0.375
-0.4375
-0.4062
-0.4219
-64141
-0.418
-0.416
-6:415
Bisection
f(xn)
33438
-5582
-14487
0.8631
-0.3137
0.2695
-0.0234
61227
09496
0.0131
f(x) x₂+1
f(x+1)
G
29.75
-12 -0.2874-44117
29.75 -0.2874-44117-0.3794 -1:2897
29.75 -6.3794-1·2897-04052-0 35 13
29.75 -04052-0.3513-04122 -0.0938
29:75 -04122-00938 -0.419
-0.0249
29.75 -0414-0·02491-0·41451-0.0066
50, approximate root.
is -0.41.
And false position method converges
faster than
method.
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