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📝 Newton's method for approximate roots (21 MCQs)

📖 From Calculus • 5. The derivative in Graphing and Applications • 21 questions available

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What is Newton's method for approximate roots?

Definition:
Newton's Method iteratively approximates roots of $f(x)=0$ using $x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$ . Starting with an initial guess $x_0$ , each iteration refines the estimate by following the tangent line to the x-axis, converging rapidly to the root.

Example:
To solve $x^2 - 2 = 0$ , start with $x_0=1$ . $x_1 = 1 - \frac{-1}{2} = 1.5$ . $x_2 = 1.5 - \frac{0.25}{3} \approx 1.4167$ , close to $\sqrt{2}$ .

Reason:
This numerical method provides efficient high-precision approximations for equations lacking algebraic solutions, leveraging local linearization via derivatives.

Easy

Medium

Hard

📝 All Newton's method for approximate roots MCQs

Q1. What is the Newton iteration formula for solving $f(x)=0$ ?

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

x_{n+1}=x_n - \dfrac{f(x_n)}{f&#039;(x_n)}

✅

x_{n+1}=x_n + \dfrac{f(x_n)}{f&#039;(x_n)}

\displaystyle x_{n+1}= \dfrac{f(x_n)}{f&#039;(x_n)}

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: The Newton iteration replaces the current estimate with the point where the tangent line at that estimate meets the x‑axis. Algebraically this gives $x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$ , which is the only option that matches the derived formula.

Q2. What is the iterative formula of Newton's Method for finding a root of a differentiable function $f(x)$ ?

x_{n+1}=x_n-\frac{f(x_n)}{f&#039;(x_n)}

✅

x_{n+1}=x_n-\frac{f&#039;(x_n)}{f(x_n)}

x_{n+1}=x_n+\frac{f(x_n)}{f&#039;(x_n)}

x_{n+1}=x_n-\frac{f&#039;&#039;(x_n)}{f&#039;(x_n)}

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: The Newton iteration replaces a point on the curve by the x‑coordinate of the tangent line's intersection with the x‑axis, giving the formula $x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$ . This expression directly follows from solving the linear approximation for the root, making it the standard iterative step.

Q3. Which situation guarantees that Newton's Method will not converge quadratically near a root?

A.When

f&#039;(r)\neq0

B.When

f&#039;(r)=0

✅

C.When

f&#039;&#039;(r)

is unbounded

D.When the function is not continuous at the root

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: Quadratic convergence relies on a simple root where the derivative at the root is non‑zero. If $f'(r)=0$ , the denominator in the Newton step becomes zero or extremely small, destroying the quadratic error reduction and causing at best linear or no convergence.

Q4. If a function $f$ has a simple root at $r$ but the initial guess $x_0$ satisfies $f'(x_0)=0$ , what is the most likely outcome of the first Newton iteration?

A.The iteration stops because division by zero occurs. ✅

B.The next iterate is undefined, causing algorithm failure.

C.The method will still converge after skipping the division.

D.The iteration produces the same value

x_0

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: Newton's formula requires division by $f'(x_n)$ . When the derivative at the initial guess is zero, the fraction $\frac{f(x_0)}{f'(x_0)}$ is undefined, so the algorithm cannot produce a new iterate and must halt, indicating a failure at the first step.

Q5. Consider $f(x)=\tan x - x$ . Near the root at $x=0$ , why might Newton's method produce large jumps for an initial guess $x_0=1.5$ ?

A.Because

f&#039;(x)

is small near 1.5, leading to a large correction term. ✅

B.Because

\tan x

has a vertical asymptote near

x=1.57

, making the derivative huge.

C.Because the function is odd, causing symmetry issues.

D.Because the iteration formula simplifies to

x_{n+1}=x_n

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: At $x_0=1.5$ the derivative $f'(x)=\sec^2 x -1$ is close to zero, so the Newton correction $-\frac{f(x_0)}{f'(x_0)}$ becomes large, causing the iterate to jump far away from the root. Small denominators amplify the step size dramatically.

Q6. Compare the convergence behavior of Newton's method applied to $f(x)=x^3$ versus $g(x)=x^3 - 0.001$ when starting at $x_0=0.1$ .

A.Both converge quadratically to the same root.

f

converges faster because its derivative at the root is zero.

g

converges quadratically while

f

stalls due to zero derivative at the root. ✅

D.Neither converges because the initial guess is too small.

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: For $f(x)=x^3$ the root at 0 has multiplicity three, making $f'(0)=0$ ; Newton's method reduces to $x_{n+1}= \frac{2}{3}x_n$ , yielding only linear convergence. The perturbed function $g$ has a simple root near 0.1, so the method retains its usual quadratic speed.

Q7. Which principle explains why Newton's method may converge to a different root than the one closest to the initial guess?

A.Mean Value Theorem

B.Intermediate Value Theorem

C.Basin of attraction concept ✅

D.Rolle's Theorem

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: Each root possesses a region of initial values that are attracted to it under iteration, known as its basin of attraction. An initial guess that lies outside the basin of the nearest root may be drawn into another basin, causing convergence to a more distant root.

Q8. For the function $f(x)=\ln(x)-1$ , if the initial guess is $x_0=0.5$ , what does the sign of $f'(x_0)$ indicate about the direction of the Newton step?

A.Positive derivative means the next iterate will be larger than

x_0

B.Negative derivative means the next iterate will be smaller than

x_0

C.Positive derivative means the correction term is negative, moving left. ✅

D.Derivative sign has no effect on the step direction.

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: Here $f'(x)=1/x$ is positive at $x_0=0.5$ . The Newton step is $-\frac{f(x_0)}{f'(x_0)}$ ; because $f(x_0)<0$ , the numerator is negative, making the whole correction positive, which moves the iterate to the right. Thus a positive derivative leads to a negative correction term.

Q9. Evaluate the impact of using a poor initial guess on the order of convergence for $f(x)=\cos x - x$ . Which statement best captures this effect?

A.The convergence remains quadratic regardless of the starting point.

B.The order degrades to linear if the guess is far from the root. ✅

C.The method may diverge, losing any convergence order.

D.The order improves to cubic when the guess is far.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: When the initial guess is distant, the linear approximation used in Newton's method may be poor, causing the error reduction factor to approach a constant rather than squaring each step. Consequently, the effective order drops from quadratic to linear, though the method can still converge if the iterates eventually enter a region where the approximation improves.

Q10. Explain why the presence of a multiple root (multiplicity >1) reduces the convergence rate of Newton's method from quadratic to linear.

A.Because the derivative at the root is zero, causing division by a small number.

B.Because the second derivative dominates the iteration formula.

C.Because the algorithm effectively becomes the secant method.

D.Because the error term no longer squares each iteration. ✅

💡 Difficulty: medium | ✅ Correct: D

📖 Explanation: For a root of multiplicity $m>1$ , the Taylor expansion shows the error after one iteration satisfies $e_{n+1}\approx\frac{m-1}{m}e_n$ . The factor $\frac{m-1}{m}$ is constant, so the error decreases linearly rather than being squared, eliminating the characteristic quadratic speed.

Q11. Given $f(x)=x^{1/3}$ , why does Newton's method fail to converge when starting at $x_0=0.1$ ?

A.The derivative

f&#039;(x)=\frac{1}{3}x^{-2/3}

becomes unbounded near zero, causing large steps.

B.The function is not differentiable at the root, violating the method's assumptions.

C.Both A and B. ✅

D.Neither; the method actually converges.

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: The root at $x=0$ is not simple; $f'(x)$ blows up as $x\to0$ , making the Newton correction $-\frac{f(x)}{f'(x)}$ extremely large and sending iterates away from the root. Simultaneously, the function lacks a well‑defined derivative at the root itself, so the method’s assumptions are breached.

Q12. Compare the behavior of Newton's method applied to $f(x)=e^x - 2$ and to $h(x)=e^x - 2x$ with the same initial guess $x_0=0$ . Which statement is accurate?

A.Both converge quadratically to the same root.

f

converges quadratically while

h

may exhibit slower convergence due to the additional linear term. ✅

h

converges faster because its derivative is larger.

D.Neither converges because the initial guess is a stationary point.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: For $f(x)=e^x-2$ the root is simple and Newton's method retains quadratic convergence. In $h(x)=e^x-2x$ the extra linear term modifies the derivative to $e^x-2$ , which is zero at the root, reducing the convergence speed and potentially causing near‑linear behavior if the initial guess is not close enough.

Q13. Synthesize how the choice of stopping criterion (e.g., $|f(x_n)|<\epsilon$ vs. $|x_{n+1}-x_n|<\epsilon$ ) can affect the perceived difficulties of Newton's method on functions with steep gradients.

A.A tolerance on

|f(x_n)|

may stop early, hiding divergence. ✅

B.A tolerance on successive iterates ensures accurate root approximation regardless of gradient.

C.Both criteria are equivalent for all functions.

D.The gradient does not influence stopping criteria.

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: When gradients are steep, $|f(x_n)|$ can become small even if the iterate is still far from the true root, causing premature termination and a false impression of success. Using the change in iterates as a criterion guards against this by requiring the actual step size to shrink, revealing any lingering divergence.

Q14. If Newton's method applied to $f(x)=x^3-2x+2$ with $x_0=1$ produces iterates that oscillate between two values, what does this suggest about the nature of the root?

A.The root is complex, so real iterates cannot converge.

B.The derivative changes sign, causing a cycle.

C.The method is stuck in a 2-cycle due to a saddle point. ✅

D.There is no real root near the initial guess.

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: A two‑cycle indicates that the iteration function has a periodic orbit of period two, which typically arises when the derivative at a nearby point has magnitude near $-1$ . This behavior reflects a saddle‑type situation where the Newton map does not settle at a fixed point, preventing convergence to a real root.

Q15. Assess why the modification $x_{n+1}=x_n-\frac{2f(x_n)}{f'(x_n)}$ (the so‑called modified Newton's method) can improve convergence for functions with a double root.

A.It doubles the correction, moving faster toward the root.

B.It effectively reduces the multiplicity, restoring quadratic convergence. ✅

C.It changes the function to its square, eliminating the root.

D.It has no effect on convergence rate.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: For a double root, the standard Newton step reduces to linear convergence because the error term contains a factor of $\frac{1}{2}$ . Multiplying the correction by two compensates for the multiplicity, effectively transforming the iteration into one that behaves like Newton's method applied to a simple root, thereby regaining quadratic speed.

Q16. Apply the concept of basin of attraction to explain why two different initial guesses $x_0= -2$ and $x_0= 2$ for $f(x)=x^3 - x$ converge to different roots.

A.Each root has its own attraction region; the guesses lie in separate basins. ✅

B.Newton's method always converges to the nearest root, regardless of basins.

C.The function symmetry forces opposite roots.

D.The method fails for symmetric functions.

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The cubic $x^3-x$ has three real roots at $-1,0,1$ . The iteration map partitions the real line into intervals (basins) that are attracted to each root. Starting at $-2$ places the iterate in the basin of $-1$ , while $2$ lies in the basin of $1$ ; thus each guess converges to a different root.

Q17. For the function $f(x)=\frac{1}{x} - \ln x$ , analyze why an initial guess $x_0=0.1$ leads to divergence, while $x_0=2$ yields rapid convergence.

A.At

x_0=0.1

the derivative

f&#039;(x)

is negative and large, causing overshoot.

B.Near zero the term

1/x

dominates, making the iteration step huge and moving away from the root.

C.The function has a vertical asymptote at

x=0

, preventing convergence from the left side.

D.Both B and C. ✅

💡 Difficulty: hard | ✅ Correct: D

📖 Explanation: When $x$ is close to zero, $1/x$ grows without bound, and the derivative $f'(x) = -\frac{1}{x^2}-\frac{1}{x}$ becomes large in magnitude, so the Newton correction $-\frac{f(x)}{f'(x)}$ is huge, propelling the iterate away from the root. Starting at $x_0=2$ avoids this region, giving a well‑behaved step and fast convergence.

Q18. Derive the order of convergence for Newton's method applied to $f(x)=\sin x$ near the root $r=0$ and explain why it differs from the typical quadratic order.

A.The derivative at the root is zero, reducing order to linear.

B.The second derivative vanishes, leading to cubic convergence.

C.The function is odd, causing the error term to be proportional to the cube of the previous error, yielding cubic convergence. ✅

D.The method fails because the root is not simple.

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: Near $r=0$ , $\sin x = x - x^3/6 + \dots$ . Substituting into Newton's formula gives $x_{n+1}=x_n - \frac{x_n - x_n^3/6}{1 - x_n^2/2}\approx \frac{x_n^3}{3}$ . Hence the new error is proportional to the cube of the previous error, indicating cubic (order‑3) convergence rather than the usual quadratic rate.

Q19. Synthesize how the presence of inflection points close to a simple root can cause Newton's method to exhibit chaotic behavior, and propose a strategy to mitigate this issue.

A.Use a smaller step size by damping the iteration.

B.Switch to the Secant method near the inflection point.

C.Choose an initial guess on the opposite side of the inflection point.

D.All of the above. ✅

💡 Difficulty: hard | ✅ Correct: D

📖 Explanation: An inflection point near a root makes the derivative change rapidly, so the Newton correction can overshoot and bounce unpredictably, producing chaotic sequences. Damping (multiplying the correction by a factor <1) reduces step size, the Secant method avoids explicit derivative calculations, and selecting a guess away from the inflection region all help stabilize convergence.

Q20. Compare the performance of Newton's method versus the Secant method for solving $f(x)=x^5 - x - 1$ in terms of iteration count and sensitivity to initial guesses.

A.Newton's method always requires fewer iterations but is more sensitive to the initial guess. ✅

B.The Secant method converges faster and is less sensitive.

C.Both methods have identical performance for this polynomial.

D.Newton's method fails entirely for quintic equations.

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: Newton's method enjoys quadratic convergence, so once the iterate is sufficiently close, it typically needs fewer steps than the Secant method, which is only super‑linear. However, Newton's method requires a good initial guess and a non‑zero derivative; the Secant method tolerates poorer guesses because it approximates the derivative, making it less sensitive.

Q21. If during Newton iteration the sequence $\{x_n\}$ approaches a cycle of period three, what does the Sharkovskii theorem imply about the existence of other periodic cycles for the underlying iteration function?

A.Existence of a period‑three cycle guarantees cycles of all other periods. ✅

B.Only cycles of periods that are powers of two exist.

C.No other cycles can exist if a period‑three cycle is present.

D.The theorem does not apply to Newton's method.

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: Sharkovskii's theorem states that the existence of a period‑three orbit for a continuous map on an interval forces the existence of periodic orbits of every possible period. Thus, finding a three‑cycle in the Newton iteration function implies that cycles of period four, five, and so on must also exist, indicating complex dynamics.

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