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Approximate a cubic root using Newton's method

Question 1 of 8

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Apply Newton’s recurrence and keep the target root bracket in view

f(x)=-2x^{3}-6x^{2}-10x-8

I=[-4,-1],\qquad x_0=-3

The target is the unique root in I. Carry unrounded values between steps: displayed decimals are terminal roundings, never the next input. Use half-away-from-zero rounding

Calculate one Newton step exactly

(a)3 marks

Enter every coefficient in descending powers, including zero for an absent term

f'(x)=A_2x^{2}+A_1x+A_0

\(\text{coefficient of }x^{2}\) \(\text{coefficient of }x\) \(\text{constant coefficient}\)

(b)1 mark

Which recurrence uses the current tangent to update the estimate?

x_{n+1}=x_n-\frac{f(x_n)}{f'(x_0)}\quad\text{for every }n

x_{n+1}=x_n+\frac{f(x_n)}{f'(x_n)}

x_{n+1}=\frac{f(x_n)}{f'(x_n)}

x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}

Enter the exact function value, tangent gradient and next estimate. Fractions are accepted

\(f(x_0)\) \(f'(x_0)\) \(x_1\)

(d)1 mark

What is x₁ geometrically?

The exact root of the cubic in every case

The y-intercept of that tangent

The derivative value at the initial input

The x-intercept of the tangent at (x₀, f(x₀))