Newton's method someone please help me with this problem :( 10 pointss!!
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Newton's method someone please help me with this problem :( 10 pointss!!

[From: ] [author: ] [Date: 12-04-10] [Hit: ]
-You need to provide an initial estimate x[1].Descartes Rule of Signs tells us there are no positive real roots, so start looking at, say,The Intermediate Value Theorem tells us that there is at least one x-value between -2 and -1 for which f(x) = 0.x[2] = -1.......
Use Newton's method to find the real root function, accurate to five decimal places

f(x) = x^5+2x^2+3


How would I go about solving this?


I used x[n] +1 = x[n] -(f(x[n]))/(f '(x[n]))

So i got f(x)=x^5 +2x^2 +3
f '(x)= 5x^4 +4x


Then I plugged it in and got x[n]=(x^5+2x^2+3)/(5x^4+4x)


Now I'm stuck what do I do next?

Can someone walk me through the steps to the answer?
Thank youuu in advance!!!!

-
You need to provide an initial estimate x[1].

Descartes' Rule of Signs tells us there are no positive real roots, so start looking at, say, negative integers (or graph the function and look for a zero)

f(-1) = 4
f(-2) = -13

The Intermediate Value Theorem tells us that there is at least one x-value between -2 and -1 for which f(x) = 0. Because |f(-1)| < |f(-2)|, let's choose x[1] = -1.3

x[2] = -1.3 - ((-1.3)^5 + 2(-1.3)² + 3) / (5(-1.3)^4 + 4(-1.3)) ≈ -1.593714
x[3] = -1.593714 - ((-1.593714)^5 + 2(-1.593714)² + 3) / (5(-1.593714)^4 + 4(-1.593714)) ≈ -1.5086506
x[4] ≈ -1.49540216
x[5] ≈ -1.495106542
x[6] ≈ -1.495106398

x[5] and x[6] agree to 6 decimal places (x[6] is in fact correct to all decimals shown) so to 5 decimal places, the root is -1.49511.

Graphing would have led me to choose x[1] = -1.5, which would have converged in fewer iterations.
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