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10 questions from this quiz
An iterative numerical technique
xn+1=xn−f(xn)f′(xn)x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}xn+1=xn−f′(xn)f(xn)
The derivative of f(x)f(x)f(x)
The tangent line to approximate the function
Choose an initial guess x0x_0x0
∣xn+1−xn∣<ϵ|x_{n+1} - x_n| < \epsilon∣xn+1−xn∣<ϵ (tolerance)
2.12.12.1
When initial guess is close to actual root
When f′(x)f'(x)f′(x) is close to zero
x≈2.0946x \approx 2.0946x≈2.0946
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