Calculus: Early Transcendentals, Chapter 4, 4.8, Section 4.8, Problem 24

x^5-3x^4+x^3-x^2-x+6=0
To solve this, using Newton's method, apply the formula:

x_(n+1)=x_n- (f(x_n))/(f'(x_n))
Let the function be:

f(x) =x^5-3x^4+x^3-x^2-x+6
Then, take its derivative.
f'(x) =5x^4-12x^3+3x^2-2x-1

Then, plug-in f(x) and f'(x) to the formula.
x_(n+1) = x_n - (x_n^5-3x_n^4+x_n^3-x_n^2-x_n+6)/(5x_n^4-12x_n^3+3x_n^2-2x_n-1)

To get the initial value of x, refer to the graph of the function. (See attached figure.)
Notice that curve has three x values when f(x)=0. These are near x=-1, x=1 and x=3. These will be our initial values for x.
Let's consider solve for the first zero.
x_1=-1
x_(2) = x_1 - (x_1^5-3x_1^4+x_1^3-x_1^2-x_1+6)/(5x_1^4-12x_1^3+3x_1^2-2x_1-1) =-1.047619047
x_3= x_2 - (x_2^5-3x_2^4+x_2^3-x_2^2-x_2+6)/(5x_2^4-12x_2^3+3x_2^2-2x_2-1) =-1.044517237
x_4 = x_3 - (x_3^5-3x_3^4+x_3^3-x_3^2-x_3+6)/(5x_3^4-12x_3^3+3x_3^2-2x_3-1)=-1.044503071
x_5 = x_4 - (x_4^5-3x_4^4+x_4^3-x_4^2-x_4+6)/(5x_4^4-12x_4^3+3x_4^2-2x_4-1)=-1.044503071
Now, we have two approximations that have same eight decimal places. Thus, one of the solution is approximately x=-1.04450307 .
Next, let's solve for the second zero.
x_1 = 1
x_2 = x_1 - (x_1^5-3x_1^4+x_1^3-x_1^2-x_1+6)/(5x_1^4-12x_1^3+3x_1^2-2x_1-1)=1.428571428
x_3 = x_2 - (x_2^5-3x_2^4+x_2^3-x_2^2-x_2+6)/(5x_2^4-12x_2^3+3x_2^2-2x_2-1)=1.336197712
x_4 = x_3 - (x_3^5-3x_3^4+x_3^3-x_3^2-x_3+6)/(5x_3^4-12x_3^3+3x_3^2-2x_3-1) =1.332589426
x_5 = x_4 - (x_4^5-3x_4^4+x_4^3-x_4^2-x_4+6)/(5x_4^4-12x_4^3+3x_4^2-2x_4-1)=1.332583155
x_6 = x_5 - (x_5^5-3x_5^4+x_5^3-x_5^2-x_5+6)/(5x_5^4-12x_5^3+3x_5^2-2x_4-1)=1.332583155
Now that we have two approximations that agree with eight decimal places, then one of the solution is approximately x=1.332583155.
Next, let's solve for the third zero.
x_1 = 3
x_2 = x_1 - (x_1^5-3x_1^4+x_1^3-x_1^2-x_1+6)/(5x_1^4-12x_1^3+3x_1^2-2x_1-1) =2.792079207
x_3 = x_2 - (x_2^5-3x_2^4+x_2^3-x_2^2-x_2+6)/(5x_2^4-12x_2^3+3x_2^2-2x_2-1)=2.715701083
x_4 = x_3 - (x_3^5-3x_3^4+x_3^3-x_3^2-x_3+6)/(5x_3^4-12x_3^3+3x_3^2-2x_3-1) =2.705675033
x_5 = x_4 - (x_4^5-3x_4^4+x_4^3-x_4^2-x_4+6)/(5x_4^4-12x_4^3+3x_4^2-2x_4-1) =2.705512135
x_6 = x_5 - (x_5^5-3x_5^4+x_5^3-x_5^2-x_5+6)/(5x_5^4-12x_5^3+3x_5^2-2x_4-1)=2.705512093
x_7 = x_6 - (x_6^5-3x_6^4+x_6^3-x_6^2-x_6+6)/(5x_6^4-12x_6^3+3x_6^2-2x_6-1)=2.705512093
Now, we have two approximations that have same eight decimal places. Thus, one of the solution is approximately x=2.705512093.
Therefore, the approximate solution ofx^5-3x^4+x^3-x^2-x+6=0 are
x={-1.04450307 ,1.332583155,2.705512093}.