In which interval does the root of equation x^3-2x-5=0 lie? Explain how to solve it.
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heya root will lie in the interval [2,3]
u can find it by putting values
let f(X) = x^3 - 2x -5 f'(X) = 3x^2-2
so , by Newton raphson method
X n+1 = Xn - f(Xn ) / f'(Xn)
here for first iteration n= 0
and Xo= 2
so X1= Xo -f(Xo)/f'(Xo)
thus X1= 2-(2^3-2*2-5)/(3*2^2-2)
X1= 2-(-1)/10
X1= 2+0.1= 2.1
u can find it by putting values
let f(X) = x^3 - 2x -5 f'(X) = 3x^2-2
so , by Newton raphson method
X n+1 = Xn - f(Xn ) / f'(Xn)
here for first iteration n= 0
and Xo= 2
so X1= Xo -f(Xo)/f'(Xo)
thus X1= 2-(2^3-2*2-5)/(3*2^2-2)
X1= 2-(-1)/10
X1= 2+0.1= 2.1
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