Description
Let f(x)=x^4-x^3. Show that the equation f(x)=75 has a solution in the interval [3,4] and use Newton’s method to find it. Please show work.
Explanation & Answer
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We have the function as
f(X)=x^4-x^3
We have x^4-x^3=75
x^4-x^3-75=0
Let g(X)=x^4-x^3-75
g(3)=81-27-75=-21
g(4)=4^4-4^3-75=117
As the g(3) value is negative and the g(4) value is positive it implies that the function has a solution in the interval in the [3,4] as the graph crosses the x-axis in this interval.
Newtons method to find the solution;
x(n+1)=xn- f(xn)/f'(xn)
Here n=3
x(n+1)=xn -(xn^4-x^n3-75)/(4xn^3-3xn^2)
Let x_0 = 3.5...recursive relation is ( 3 x^4 - 2 x³ + 75 ) / ( 4 x³ - 3 x²)
Using a calculator 5 times the 10 place accuracy...3.2285
I hope this would help you,but if you have any doubt regarding this,you can surely ask. :)Review
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