bmte-144-sample-solution-00121e82-06b0-4090-b5ab-7e9d6642f92d
a) Find the largest real root
α
α
alpha \alpha α
f
(
x
)
=
x
6
−
x
−
1
=
0
f
(
x
)
=
x
6
−
x
−
1
=
0
f(x)=x^(6)-x-1=0 f(x)=x^6-x-1=0 f ( x ) = x 6 − x − 1 = 0
(
x
0
=
2
,
x
1
=
1
)
x
0
=
2
,
x
1
=
1
(x_(0)=2,x_(1)=1) \left(x_0=2, x_1=1\right) ( x 0 = 2 , x 1 = 1 )
Answer:
i) Bisection method
Here
x
6
−
x
−
1
=
0
x
6
−
x
−
1
=
0
x^(6)-x-1=0 x^6-x-1=0 x 6 − x − 1 = 0
Let
f
(
x
)
=
x
6
−
x
−
1
f
(
x
)
=
x
6
−
x
−
1
f(x)=x^(6)-x-1 f(x)=x^6-x-1 f ( x ) = x 6 − x − 1
1
st
1
st
1^(“st “) 1^{\text {st }} 1 st iteration :
Here
f
(
1
)
=
−
1
<
0
f
(
1
)
=
−
1
<
0
f(1)=-1 < 0 f(1)=-1<0 f ( 1 ) = − 1 < 0 and
f
(
2
)
=
61
>
0
f
(
2
)
=
61
>
0
f(2)=61 > 0 f(2)=61>0 f ( 2 ) = 61 > 0
∴
∴
:. \therefore ∴ Now, Root lies between 1 and 2
x
0
=
1
+
2
2
=
1.5
x
0
=
1
+
2
2
=
1.5
x_(0)=(1+2)/(2)=1.5 x_0=\frac{1+2}{2}=1.5 x 0 = 1 + 2 2 = 1.5
f
(
x
0
)
=
f
(
1.5
)
=
1.5
6
−
1.5
−
1
=
8.8906
>
0
f
x
0
=
f
(
1.5
)
=
1.5
6
−
1.5
−
1
=
8.8906
>
0
f(x_(0))=f(1.5)=1.5^(6)-1.5-1=8.8906 > 0 f\left(x_0\right)=f(1.5)=1.5^6-1.5-1=8.8906>0 f ( x 0 ) = f ( 1.5 ) = 1.5 6 − 1.5 − 1 = 8.8906 > 0
2
nd
2
nd
2^(“nd “) 2^{\text {nd }} 2 nd iteration :
Here
f
(
1
)
=
−
1
<
0
f
(
1
)
=
−
1
<
0
f(1)=-1 < 0 f(1)=-1<0 f ( 1 ) = − 1 < 0 and
f
(
1.5
)
=
8.8906
>
0
f
(
1.5
)
=
8.8906
>
0
f(1.5)=8.8906 > 0 f(1.5)=8.8906>0 f ( 1.5 ) = 8.8906 > 0
∴
∴
:. \therefore ∴ Now, Root lies between 1 and 1.5
x
1
=
1
+
1.5
2
=
1.25
x
1
=
1
+
1.5
2
=
1.25
x_(1)=(1+1.5)/(2)=1.25 x_1=\frac{1+1.5}{2}=1.25 x 1 = 1 + 1.5 2 = 1.25
f
(
x
1
)
=
f
(
1.25
)
=
1.25
6
−
1.25
−
1
=
1.5647
>
0
f
x
1
=
f
(
1.25
)
=
1.25
6
−
1.25
−
1
=
1.5647
>
0
f(x_(1))=f(1.25)=1.25^(6)-1.25-1=1.5647 > 0 f\left(x_1\right)=f(1.25)=1.25^6-1.25-1=1.5647>0 f ( x 1 ) = f ( 1.25 ) = 1.25 6 − 1.25 − 1 = 1.5647 > 0
3
r
d
3
r
d
3^(rd) 3^{r d} 3 r d iteration :
Here
f
(
1
)
=
−
1
<
0
f
(
1
)
=
−
1
<
0
f(1)=-1 < 0 f(1)=-1<0 f ( 1 ) = − 1 < 0 and
f
(
1.25
)
=
1.5647
>
0
f
(
1.25
)
=
1.5647
>
0
f(1.25)=1.5647 > 0 f(1.25)=1.5647>0 f ( 1.25 ) = 1.5647 > 0
∴
∴
:. \therefore ∴ Now, Root lies between 1 and 1.25
x
2
=
1
+
1.25
2
=
1.125
x
2
=
1
+
1.25
2
=
1.125
x_(2)=(1+1.25)/(2)=1.125 x_2=\frac{1+1.25}{2}=1.125 x 2 = 1 + 1.25 2 = 1.125
f
(
x
2
)
=
f
(
1.125
)
=
1.125
6
−
1.125
−
1
=
−
0.0977
<
0
f
x
2
=
f
(
1.125
)
=
1.125
6
−
1.125
−
1
=
−
0.0977
<
0
f(x_(2))=f(1.125)=1.125^(6)-1.125-1=-0.0977 < 0 f\left(x_2\right)=f(1.125)=1.125^6-1.125-1=-0.0977<0 f ( x 2 ) = f ( 1.125 ) = 1.125 6 − 1.125 − 1 = − 0.0977 < 0 After three iterations, the interval containing the largest real root
α
α
alpha \alpha α is narrowed down to
[
1.125
,
1.25
]
[
1.125
,
1.25
]
[1.125,1.25] [1.125,1.25] [ 1.125 , 1.25 ] . The midpoint of this interval,
x
3
=
1.1875
x
3
=
1.1875
x_(3)=1.1875 x_3=1.1875 x 3 = 1.1875 , is the best approximation of the root after three iterations using the bisection method.
ii) Secant method
(
x
0
=
2
,
x
1
=
1
)
x
0
=
2
,
x
1
=
1
(x_(0)=2,x_(1)=1) \left(x_0=2, x_1=1\right) ( x 0 = 2 , x 1 = 1 ) .
Let
f
(
x
)
=
x
6
−
x
−
1
f
(
x
)
=
x
6
−
x
−
1
f(x)=x^(6)-x-1 f(x)=x^6-x-1 f ( x ) = x 6 − x − 1
1
st
1
st
1^(“st “) 1^{\text {st }} 1 st iteration :
x
0
=
1
x
0
=
1
x_(0)=1 x_0=1 x 0 = 1 and
x
1
=
2
x
1
=
2
x_(1)=2 x_1=2 x 1 = 2
f
(
x
0
)
=
f
(
1
)
=
−
1
f
x
0
=
f
(
1
)
=
−
1
f(x_(0))=f(1)=-1 f\left(x_0\right)=f(1)=-1 f ( x 0 ) = f ( 1 ) = − 1 and
f
(
x
1
)
=
f
(
2
)
=
61
f
x
1
=
f
(
2
)
=
61
f(x_(1))=f(2)=61 f\left(x_1\right)=f(2)=61 f ( x 1 ) = f ( 2 ) = 61
∴
x
2
=
x
0
−
f
(
x
0
)
⋅
x
1
−
x
0
f
(
x
1
)
−
f
(
x
0
)
∴
x
2
=
x
0
−
f
x
0
⋅
x
1
−
x
0
f
x
1
−
f
x
0
:.x_(2)=x_(0)-f(x_(0))*(x_(1)-x_(0))/(f(x_(1))-f(x_(0))) \therefore x_2=x_0-f\left(x_0\right) \cdot \frac{x_1-x_0}{f\left(x_1\right)-f\left(x_0\right)} ∴ x 2 = x 0 − f ( x 0 ) ⋅ x 1 − x 0 f ( x 1 ) − f ( x 0 )
x
2
=
1
−
(
−
1
)
⋅
2
−
1
61
−
(
−
1
)
x
2
=
1
−
(
−
1
)
⋅
2
−
1
61
−
(
−
1
)
x_(2)=1-(-1)*(2-1)/(61-(-1)) x_2=1-(-1) \cdot \frac{2-1}{61-(-1)} x 2 = 1 − ( − 1 ) ⋅ 2 − 1 61 − ( − 1 )
x
2
=
1.0161
x
2
=
1.0161
x_(2)=1.0161 x_2=1.0161 x 2 = 1.0161
∴
f
(
x
2
)
=
f
(
1.0161
)
=
1.0161
6
−
1.0161
−
1
=
−
0.9154
∴
f
x
2
=
f
(
1.0161
)
=
1.0161
6
−
1.0161
−
1
=
−
0.9154
:.f(x_(2))=f(1.0161)=1.0161^(6)-1.0161-1=-0.9154 \therefore f\left(x_2\right)=f(1.0161)=1.0161^6-1.0161-1=-0.9154 ∴ f ( x 2 ) = f ( 1.0161 ) = 1.0161 6 − 1.0161 − 1 = − 0.9154
2
nd
2
nd
2^(“nd “) 2^{\text {nd }} 2 nd iteration :
x
1
=
2
x
1
=
2
x_(1)=2 x_1=2 x 1 = 2 and
x
2
=
1.0161
x
2
=
1.0161
x_(2)=1.0161 x_2=1.0161 x 2 = 1.0161
f
(
x
1
)
=
f
(
2
)
=
61
and
f
(
x
2
)
=
f
(
1.0161
)
=
−
0.9154
f
x
1
=
f
(
2
)
=
61
and
f
x
2
=
f
(
1.0161
)
=
−
0.9154
f(x_(1))=f(2)=61″ and “f(x_(2))=f(1.0161)=-0.9154 f\left(x_1\right)=f(2)=61 \text { and } f\left(x_2\right)=f(1.0161)=-0.9154 f ( x 1 ) = f ( 2 ) = 61 and f ( x 2 ) = f ( 1.0161 ) = − 0.9154
∴
x
3
=
x
1
−
f
(
x
1
)
⋅
x
2
−
x
1
f
(
x
2
)
−
f
(
x
1
)
∴
x
3
=
x
1
−
f
x
1
⋅
x
2
−
x
1
f
x
2
−
f
x
1
:.x_(3)=x_(1)-f(x_(1))*(x_(2)-x_(1))/(f(x_(2))-f(x_(1))) \therefore x_3=x_1-f\left(x_1\right) \cdot \frac{x_2-x_1}{f\left(x_2\right)-f\left(x_1\right)} ∴ x 3 = x 1 − f ( x 1 ) ⋅ x 2 − x 1 f ( x 2 ) − f ( x 1 )
x
3
=
2
−
61
⋅
1.0161
−
2
−
0.9154
−
61
x
3
=
2
−
61
⋅
1.0161
−
2
−
0.9154
−
61
x_(3)=2-61*(1.0161-2)/(-0.9154-61) x_3=2-61 \cdot \frac{1.0161-2}{-0.9154-61} x 3 = 2 − 61 ⋅ 1.0161 − 2 − 0.9154 − 61
x
3
=
1.0307
x
3
=
1.0307
x_(3)=1.0307 x_3=1.0307 x 3 = 1.0307
∴
f
(
x
3
)
=
f
(
1.0307
)
=
1.0307
6
−
1.0307
−
1
=
−
0.8319
∴
f
x
3
=
f
(
1.0307
)
=
1.0307
6
−
1.0307
−
1
=
−
0.8319
:.f(x_(3))=f(1.0307)=1.0307^(6)-1.0307-1=-0.8319 \therefore f\left(x_3\right)=f(1.0307)=1.0307^6-1.0307-1=-0.8319 ∴ f ( x 3 ) = f ( 1.0307 ) = 1.0307 6 − 1.0307 − 1 = − 0.8319
3
r
d
3
r
d
3^(rd) 3^{r d} 3 r d iteration :
x
2
=
1.0161
and
x
3
=
1.0307
x
2
=
1.0161
and
x
3
=
1.0307
x_(2)=1.0161″ and “x_(3)=1.0307 x_2=1.0161 \text { and } x_3=1.0307 x 2 = 1.0161 and x 3 = 1.0307
f
(
x
2
)
=
f
(
1.0161
)
=
−
0.9154
and
f
(
x
3
)
=
f
(
1.0307
)
=
−
0.8319
f
x
2
=
f
(
1.0161
)
=
−
0.9154
and
f
x
3
=
f
(
1.0307
)
=
−
0.8319
f(x_(2))=f(1.0161)=-0.9154″ and “f(x_(3))=f(1.0307)=-0.8319 f\left(x_2\right)=f(1.0161)=-0.9154 \text { and } f\left(x_3\right)=f(1.0307)=-0.8319 f ( x 2 ) = f ( 1.0161 ) = − 0.9154 and f ( x 3 ) = f ( 1.0307 ) = − 0.8319
∴
x
4
=
x
2
−
f
(
x
2
)
⋅
x
3
−
x
2
f
(
x
3
)
−
f
(
x
2
)
x
4
=
1.0161
−
(
−
0.9154
)
⋅
1.0307
−
1.0161
−
0.8319
−
(
−
0.9154
)
x
4
=
1.1757
∴
f
(
x
4
)
=
f
(
1.1757
)
=
1.1757
6
−
1.1757
−
1
=
0.4652
∴
x
4
=
x
2
−
f
x
2
⋅
x
3
−
x
2
f
x
3
−
f
x
2
x
4
=
1.0161
−
(
−
0.9154
)
⋅
1.0307
−
1.0161
−
0.8319
−
(
−
0.9154
)
x
4
=
1.1757
∴
f
x
4
=
f
(
1.1757
)
=
1.1757
6
−
1.1757
−
1
=
0.4652
{:[:.x_(4)=x_(2)-f(x_(2))*(x_(3)-x_(2))/(f(x_(3))-f(x_(2)))],[x_(4)=1.0161-(-0.9154)*(1.0307-1.0161)/(-0.8319-(-0.9154))],[x_(4)=1.1757],[:.f(x_(4))=f(1.1757)=1.1757^(6)-1.1757-1=0.4652]:} \begin{aligned}
& \therefore x_4=x_2-f\left(x_2\right) \cdot \frac{x_3-x_2}{f\left(x_3\right)-f\left(x_2\right)} \\
& x_4=1.0161-(-0.9154) \cdot \frac{1.0307-1.0161}{-0.8319-(-0.9154)} \\
& x_4=1.1757 \\
& \therefore f\left(x_4\right)=f(1.1757)=1.1757^6-1.1757-1=0.4652
\end{aligned} ∴ x 4 = x 2 − f ( x 2 ) ⋅ x 3 − x 2 f ( x 3 ) − f ( x 2 ) x 4 = 1.0161 − ( − 0.9154 ) ⋅ 1.0307 − 1.0161 − 0.8319 − ( − 0.9154 ) x 4 = 1.1757 ∴ f ( x 4 ) = f ( 1.1757 ) = 1.1757 6 − 1.1757 − 1 = 0.4652 To summarize, after three iterations of the secant method, the approximation of the largest real root
α
α
alpha \alpha α of the equation
f
(
x
)
=
x
6
−
x
−
1
=
0
f
(
x
)
=
x
6
−
x
−
1
=
0
f(x)=x^(6)-x-1=0 f(x)=x^6-x-1=0 f ( x ) = x 6 − x − 1 = 0 is
x
4
=
1.1757
x
4
=
1.1757
x_(4)=1.1757 x_4=1.1757 x 4 = 1.1757 . The value of the function at this point is
f
(
x
4
)
=
0.465365
f
x
4
=
0.465365
f(x_(4))=0.465365 f\left(x_4\right)=0.465365 f ( x 4 ) = 0.465365 , indicating that the root is close to this value.
b) Find the number of positive and negative roots of the polynomial
P
(
x
)
=
x
3
−
3
x
2
+
4
x
−
5
P
(
x
)
=
x
3
−
3
x
2
+
4
x
−
5
P(x)=x^(3)-3x^(2)+4x-5 P(x)=x^3-3 x^2+4 x-5 P ( x ) = x 3 − 3 x 2 + 4 x − 5 . Find
P
(
2
)
P
(
2
)
P(2) P(2) P ( 2 ) and
P
′
(
2
)
P
′
(
2
)
P^(‘)(2) P^{\prime}(2) P ′ ( 2 ) using synthetic division method.
Answer:
To find the number of positive and negative roots of the polynomial
P
(
x
)
=
x
3
−
3
x
2
+
4
x
−
5
P
(
x
)
=
x
3
−
3
x
2
+
4
x
−
5
P(x)=x^(3)-3x^(2)+4x-5 P(x) = x^3 – 3x^2 + 4x – 5 P ( x ) = x 3 − 3 x 2 + 4 x − 5 , we can use Descartes’ Rule of Signs. This rule states that the number of positive real roots of a polynomial is either equal to the number of sign changes between consecutive coefficients or less than that by an even number. Similarly, the number of negative real roots is determined by the number of sign changes in the coefficients of the polynomial with each
x
x
x x x replaced by
−
x
−
x
-x -x − x .
Finding the Number of Positive Roots:
The given polynomial is
P
(
x
)
=
x
3
−
3
x
2
+
4
x
−
5
P
(
x
)
=
x
3
−
3
x
2
+
4
x
−
5
P(x)=x^(3)-3x^(2)+4x-5 P(x) = x^3 – 3x^2 + 4x – 5 P ( x ) = x 3 − 3 x 2 + 4 x − 5 . The coefficients are 1, -3, 4, -5. The sign changes in the coefficients are from 1 to -3, -3 to 4, and 4 to -5. So, there are 3 sign changes.
Number of Positive Roots: 3 or 1 (since it can be less by an even number).
Finding the Number of Negative Roots:
Replace
x
x
x x x with
−
x
−
x
-x -x − x in
P
(
x
)
P
(
x
)
P(x) P(x) P ( x ) , we get
P
(
−
x
)
=
−
x
3
−
3
x
2
−
4
x
−
5
P
(
−
x
)
=
−
x
3
−
3
x
2
−
4
x
−
5
P(-x)=-x^(3)-3x^(2)-4x-5 P(-x) = -x^3 – 3x^2 – 4x – 5 P ( − x ) = − x 3 − 3 x 2 − 4 x − 5 . The coefficients are -1, -3, -4, -5. There are no sign changes.
Number of Negative Roots: 0 (since there are no sign changes).
Calculating
P
(
2
)
P
(
2
)
P(2) P(2) P ( 2 )
P
′
(
2
)
P
′
(
2
)
P^(‘)(2) P'(2) P ′ ( 2 )
Calculate
P
(
2
)
P
(
2
)
P(2) P(2) P ( 2 ) We use synthetic division to divide
P
(
x
)
P
(
x
)
P(x) P(x) P ( x ) by
x
−
2
x
−
2
x-2 x – 2 x − 2 .
2
1
−
3
4
−
5
2
−
2
4
1
−
1
2
−
1
2
1
−
3
4
−
5
2
−
2
4
1
−
1
2
−
1
{:[2,1,-3,4,-5],[,,2,-2,4],[,1,-1,2,-1]:} \begin{array}{c|ccc}
2 & 1 & -3 & 4 & -5 \\
\hline
& & 2 & -2 & 4 \\
\hline
& 1 & -1 & 2 & -1
\end{array} 2 1 − 3 4 − 5 2 − 2 4 1 − 1 2 − 1 The remainder is -1, so
P
(
2
)
=
−
1
P
(
2
)
=
−
1
P(2)=-1 P(2) = -1 P ( 2 ) = − 1 .
Calculate
P
′
(
2
)
P
′
(
2
)
P^(‘)(2) P'(2) P ′ ( 2 ) The derivative of a polynomial can be found by multiplying each term by its degree and reducing the degree by one. For
P
(
x
)
=
x
3
−
3
x
2
+
4
x
−
5
P
(
x
)
=
x
3
−
3
x
2
+
4
x
−
5
P(x)=x^(3)-3x^(2)+4x-5 P(x) = x^3 – 3x^2 + 4x – 5 P ( x ) = x 3 − 3 x 2 + 4 x − 5 , the derivative is
P
′
(
x
)
=
3
x
2
−
6
x
+
4
P
′
(
x
)
=
3
x
2
−
6
x
+
4
P^(‘)(x)=3x^(2)-6x+4 P'(x) = 3x^2 – 6x + 4 P ′ ( x ) = 3 x 2 − 6 x + 4 .
Now, we use synthetic division to find
P
′
(
2
)
P
′
(
2
)
P^(‘)(2) P'(2) P ′ ( 2 ) by dividing
P
′
(
x
)
P
′
(
x
)
P^(‘)(x) P'(x) P ′ ( x ) by
x
−
2
x
−
2
x-2 x – 2 x − 2 .
2
3
−
6
4
6
0
3
0
4
2
3
−
6
4
6
0
3
0
4
{:[2,3,-6,4],[,,6,0],[,3,0,4]:} \begin{array}{c|cc}
2 & 3 & -6 & 4 \\
\hline
& & 6 & 0 \\
\hline
& 3 & 0 & 4
\end{array} 2 3 − 6 4 6 0 3 0 4 The remainder is 4, so
P
′
(
2
)
=
4
P
′
(
2
)
=
4
P^(‘)(2)=4 P'(2) = 4 P ′ ( 2 ) = 4 .
To summarize, the polynomial
P
(
x
)
=
x
3
−
3
x
2
+
4
x
−
5
P
(
x
)
=
x
3
−
3
x
2
+
4
x
−
5
P(x)=x^(3)-3x^(2)+4x-5 P(x) = x^3 – 3x^2 + 4x – 5 P ( x ) = x 3 − 3 x 2 + 4 x − 5 has either 3 or 1 positive real roots and 0 negative real roots. Additionally,
P
(
2
)
=
−
1
P
(
2
)
=
−
1
P(2)=-1 P(2) = -1 P ( 2 ) = − 1 and
P
′
(
2
)
=
4
P
′
(
2
)
=
4
P^(‘)(2)=4 P'(2) = 4 P ′ ( 2 ) = 4 .
c) Solve
x
3
−
9
x
+
1
=
0
x
3
−
9
x
+
1
=
0
x^(3)-9x+1=0 x^3-9 x+1=0 x 3 − 9 x + 1 = 0 for the root lying between 2 and 4 by the method of false position. Perform two iterations.
Answer:
Let
f
(
x
)
=
x
3
−
9
x
+
1
f
(
x
)
=
x
3
−
9
x
+
1
f(x)=x^(3)-9x+1 f(x)=x^3-9 x+1 f ( x ) = x 3 − 9 x + 1
Here
x
x
x x x 2
3
4
f
(
x
)
f
(
x
)
f(x) f(x) f ( x ) -9
1
29
x 2 3 4
f(x) -9 1 29 | $x$ | 2 | 3 | 4 |
| :—: | :—: | :—: | :—: |
| $f(x)$ | -9 | 1 | 29 |
Here
f
(
2
)
=
−
9
<
0
f
(
2
)
=
−
9
<
0
f(2)=-9 < 0 f(2)=-9<0 f ( 2 ) = − 9 < 0 and
f
(
3
)
=
1
>
0
f
(
3
)
=
1
>
0
f(3)=1 > 0 f(3)=1>0 f ( 3 ) = 1 > 0
∴
∴
:. \therefore ∴ Root lies between 2 and 3
1
s
t
1
s
t
1^(st) 1^{s t} 1 s t iteration :
Here
f
(
2
)
=
−
9
<
0
f
(
2
)
=
−
9
<
0
f(2)=-9 < 0 f(2)=-9<0 f ( 2 ) = − 9 < 0 and
f
(
3
)
=
1
>
0
f
(
3
)
=
1
>
0
f(3)=1 > 0 f(3)=1>0 f ( 3 ) = 1 > 0
∴
∴
:. \therefore ∴ Now, Root lies between
x
0
=
2
x
0
=
2
x_(0)=2 x_0=2 x 0 = 2 and
x
1
=
3
x
1
=
3
x_(1)=3 x_1=3 x 1 = 3
x
2
=
x
0
−
f
(
x
0
)
⋅
x
1
−
x
0
f
(
x
1
)
−
f
(
x
0
)
x
2
=
2
−
(
−
9
)
⋅
3
−
2
1
−
(
−
9
)
x
2
=
2.9
x
2
=
x
0
−
f
x
0
⋅
x
1
−
x
0
f
x
1
−
f
x
0
x
2
=
2
−
(
−
9
)
⋅
3
−
2
1
−
(
−
9
)
x
2
=
2.9
{:[x_(2)=x_(0)-f(x_(0))*(x_(1)-x_(0))/(f(x_(1))-f(x_(0)))],[x_(2)=2-(-9)*(3-2)/(1-(-9))],[x_(2)=2.9]:} \begin{aligned}
& x_2=x_0-f\left(x_0\right) \cdot \frac{x_1-x_0}{f\left(x_1\right)-f\left(x_0\right)} \\
& x_2=2-(-9) \cdot \frac{3-2}{1-(-9)} \\
& x_2=2.9
\end{aligned} x 2 = x 0 − f ( x 0 ) ⋅ x 1 − x 0 f ( x 1 ) − f ( x 0 ) x 2 = 2 − ( − 9 ) ⋅ 3 − 2 1 − ( − 9 ) x 2 = 2.9
f
(
x
2
)
=
f
(
2.9
)
=
2.9
3
−
9
⋅
2.9
+
1
=
−
0.711
<
0
f
x
2
=
f
(
2.9
)
=
2.9
3
−
9
⋅
2.9
+
1
=
−
0.711
<
0
f(x_(2))=f(2.9)=2.9^(3)-9*2.9+1=-0.711 < 0 f\left(x_2\right)=f(2.9)=2.9^3-9 \cdot 2.9+1=-0.711<0 f ( x 2 ) = f ( 2.9 ) = 2.9 3 − 9 ⋅ 2.9 + 1 = − 0.711 < 0
2
nd
2
nd
2^(“nd “) 2^{\text {nd }} 2 nd iteration :
Here
f
(
2.9
)
=
−
0.711
<
0
f
(
2.9
)
=
−
0.711
<
0
f(2.9)=-0.711 < 0 f(2.9)=-0.711<0 f ( 2.9 ) = − 0.711 < 0 and
f
(
3
)
=
1
>
0
f
(
3
)
=
1
>
0
f(3)=1 > 0 f(3)=1>0 f ( 3 ) = 1 > 0
∴
∴
:. \therefore ∴ Now, Root lies between
x
0
=
2.9
x
0
=
2.9
x_(0)=2.9 x_0=2.9 x 0 = 2.9 and
x
1
=
3
x
1
=
3
x_(1)=3 x_1=3 x 1 = 3
x
3
=
x
0
−
f
(
x
0
)
⋅
x
1
−
x
0
f
(
x
1
)
−
f
(
x
0
)
x
3
=
2.9
−
(
−
0.711
)
⋅
3
−
2.9
1
−
(
−
0.711
)
x
3
=
2.9416
x
3
=
x
0
−
f
x
0
⋅
x
1
−
x
0
f
x
1
−
f
x
0
x
3
=
2.9
−
(
−
0.711
)
⋅
3
−
2.9
1
−
(
−
0.711
)
x
3
=
2.9416
{:[x_(3)=x_(0)-f(x_(0))*(x_(1)-x_(0))/(f(x_(1))-f(x_(0)))],[x_(3)=2.9-(-0.711)*(3-2.9)/(1-(-0.711))],[x_(3)=2.9416]:} \begin{aligned}
& x_3=x_0-f\left(x_0\right) \cdot \frac{x_1-x_0}{f\left(x_1\right)-f\left(x_0\right)} \\
& x_3=2.9-(-0.711) \cdot \frac{3-2.9}{1-(-0.711)} \\
& x_3=2.9416
\end{aligned} x 3 = x 0 − f ( x 0 ) ⋅ x 1 − x 0 f ( x 1 ) − f ( x 0 ) x 3 = 2.9 − ( − 0.711 ) ⋅ 3 − 2.9 1 − ( − 0.711 ) x 3 = 2.9416
f
(
x
3
)
=
f
(
2.9416
)
=
2.9416
3
−
9
⋅
2.9416
+
1
=
−
0.0215
<
0
f
x
3
=
f
(
2.9416
)
=
2.9416
3
−
9
⋅
2.9416
+
1
=
−
0.0215
<
0
f(x_(3))=f(2.9416)=2.9416^(3)-9*2.9416+1=-0.0215 < 0 f\left(x_3\right)=f(2.9416)=2.9416^3-9 \cdot 2.9416+1=-0.0215<0 f ( x 3 ) = f ( 2.9416 ) = 2.9416 3 − 9 ⋅ 2.9416 + 1 = − 0.0215 < 0