- Compare the midpoint's sign with each endpoint sign. - Keep the half-interval that still has opposite signs at its ends. - The magnitude of the function values is not what decides the next bracket.

1. The initial bracket \([2,3]\) has opposite endpoint signs. 2. At the midpoint, \(F(2.5)=-2<0\), which has the same sign as \(F(2)\). 3. The sign change therefore remains between \(2.5\) and \(3\), so the next interval is \([2.5,3]\).

Use \([2.5,3]\) for the next step.

- Compare the endpoints of consecutive panels. - The midpoint of one bracket becomes an endpoint of the next retained bracket. - Track how the interval width changes.

1. Panel a) shows the initial bracket \([0,4]\), whose midpoint is \(2\). 2. Panel b) shows the retained bracket \([2,4]\), so the midpoint test at \(2\) showed that the sign change was in the right half. Its midpoint is \(3\). 3. Panel c) shows the retained bracket \([2,3]\), so the midpoint test at \(3\) showed that the sign change was in the left half. 4. Each step keeps one half of the previous bracket, so the interval width is halved.

The tested midpoints are \(2\) and then \(3\). The brackets shrink from \([0,4]\) to \([2,4]\) to \([2,3]\), halving the width at each step.

- Read the two marked x-values. - What is special about having opposite signs at the endpoints of a continuous function? - State the entire bracket, not only its midpoint.

1. The number line gives \(A=1\) and \(B=3\). 2. Continuity and opposite signs at the endpoints guarantee at least one zero between them. 3. Hence \([1,3]\) is a valid bisection starting bracket.

\([1,3]\).

- Identify both ends of the highlighted interval. - Bisection divides the current bracket into two equal lengths. - Use the average of the endpoints.

1. Read the endpoints of the highlighted bracket as \(2\) and \(6\). 2. Bisection tests the midpoint \(\frac{2+6}{2}=4\).

\(x=4\).

- Compute the midpoint of the starting bracket. - Substitute that value into the function. - If the midpoint is already a root, is further halving needed?

1. The midpoint is \(m=\frac{1+3}{2}=2\). 2. Evaluate \(f(2)=2^2-4=0\). 3. The midpoint is an exact root, so the bisection process stops without choosing another half-interval.

The midpoint is \(2\), and \(f(2)=0\). Stop: \(x=2\) is an exact root.

- Start by confirming opposite signs at the two endpoints. - At each step, test the midpoint and keep the half whose endpoint values still have opposite signs. - Record the new bracket before moving to the next midpoint.

1. \(f(1)=-1<0\) and \(f(2)=5>0\), so a zero lies in \([1,2]\). 2. The first midpoint is \(1.5\), with \(f(1.5)=0.875>0\). Keep \([1,1.5]\). 3. The second midpoint is \(1.25\), with \(f(1.25)=-0.296875<0\). Keep \([1.25,1.5]\). 4. The third midpoint is \(1.375\), with \(f(1.375)=0.224609375>0\). Keep \([1.25,1.375]\).

After three midpoint tests, the zero is bracketed in \([1.25,1.375]\).

- Recall what sign condition starts the bisection method. - Check whether the absence of a sign change is only a method issue or reflects the function itself. - Rewrite the quadratic in a form that makes its minimum value clear.

1. The standard bisection guarantee requires a continuous function with opposite signs at the two endpoints. Here both endpoint values are positive. 2. Complete the square: \(g(x)=(x-2)^2+1\). 3. Since \(g(x)\ge1\) for every real \(x\), the function has no real zero. Halving the interval cannot produce a sign-changing bracket.

The bisection guarantee does not apply because the endpoint values have the same sign. In fact, \(g(x)=(x-2)^2+1>0\) for all real \(x\), so there is no real zero.

- Turn the equation into a zero-finding problem for one function. - At each midpoint, only the sign is needed to choose the next half. - Keep the half whose endpoint function values have opposite signs.

1. \(F(0)=1>0\) and \(F(1)=\cos1-1<0\), so a zero is bracketed in \([0,1]\). 2. The first midpoint is \(0.5\). Since \(F(0.5)=\cos0.5-0.5>0\), keep \([0.5,1]\). 3. The second midpoint is \(0.75\). Since \(F(0.75)=\cos0.75-0.75<0\), keep \([0.5,0.75]\).

After two bisection steps, the solution is bracketed in \([0.5,0.75]\).

- Focus on what information each method preserves after one iteration. - One method keeps two endpoints around the target; the other follows a tangent-line intercept. - Consider how division by a very small slope affects the size of a Newton step.

1. Bisection keeps an interval whose endpoint function values have opposite signs. Continuity guarantees that at least one zero remains inside every retained bracket. 2. Each bisection step halves the bracket, so the uncertainty decreases predictably. 3. Newton's update divides by the derivative. If the derivative at the current approximation is very small, the correction term can be very large, so the next approximation can move far outside the original neighborhood. 4. Thus bisection has a bracketing guarantee under its hypotheses, while Newton's method depends strongly on the starting value and tangent slope.

Bisection preserves an opposite-sign bracket and halves it at every step, so the zero remains trapped. Newton's method can jump far away when the derivative in its denominator is close to zero.

- Track the bracket width rather than any particular function value. - Each step cuts the current width in half. - Compare powers of \(2\) with the reduction factor you need.

1. After \(n\) bisection steps, the bracket width is \(\frac{8}{2^n}\). 2. Require \(\frac{8}{2^n}<0.01\), so \(2^n>800\). 3. Since \(2^9=512\) and \(2^{10}=1024\), the smallest possible value is \(n=10\).

\(10\) bisection steps.

- Compare the midpoint sign with both endpoint signs. - The retained interval must still have opposite signs at its endpoints. - Do not choose a half just because it contains the midpoint.

1. The midpoint value \(F(1.4)>0\) has the same sign as \(F(1.6)\). 2. The sign change is therefore between \(1.2\) and \(1.4\), not between \(1.4\) and \(1.6\). 3. The correct next bracket is \([1.2,1.4]\).

Wanjiku kept two endpoints with the same sign. The correct next bracket is \([1.2,1.4]\).

- Convert the horizontal-tangent condition into a zero-finding problem. - Use endpoint signs to start and preserve the bisection bracket. - Continue until the bracket determines a unique rounding to the requested precision.

1. A horizontal tangent occurs where \(g'(x)=0\). Let \(F(x)=x^3-2x-2\). 2. \(F(1)=-3<0\) and \(F(2)=2>0\), so bisection applies on \([1,2]\). 3. Repeated midpoint tests give \([1.5,2]\), \([1.75,2]\), \([1.75,1.875]\), and then progressively smaller brackets. 4. After seven bisection steps, the retained bracket is \([1.765625,1.7734375]\). Every value in this bracket rounds to \(1.77\) to two decimal places. 5. Therefore, the horizontal tangent occurs at \(x\approx1.77\).

The horizontal tangent occurs at \(x\approx1.77\).

- Read the endpoints of the final, narrower interval. - Where is its midpoint relative to the two ends? - The root remains inside the bracket, so how far away could it be from the midpoint?

1. The final bracket runs from \(3.75\) to \(5\), so its width is \(5-3.75=1.25\). 2. The midpoint is \(\frac{3.75+5}{2}=4.375\). 3. The root lies in this bracket, so it can be no farther than half its width from the midpoint. 4. The maximum possible absolute error is \(\frac{1.25}{2}=0.625\).

The absolute error is at most \(0.625\).

- Bisection retains a bracket with opposite signs at its ends. - The sign at the left endpoint is already known. - Treat a zero midpoint separately from a positive or negative midpoint.

1. The left endpoint has negative function value. 2. To keep opposite signs on \([1,3]\), the new right endpoint must have positive function value, so \(f(3)>0\). 3. If \(f(3)=0\), the midpoint is an exact root and the method stops instead of choosing a new bracket.

\(f(3)>0\) is needed to retain \([1,3]\). If \(f(3)=0\), stop because \(3\) is a root.

- Besides opposite endpoint signs, what does the bisection theorem assume? - Is the formula defined at every input in the proposed bracket? - What would distinguish a genuine zero from a discontinuity?

1. The endpoint values are \(f(0)=-1\) and \(f(2)=1\), so their signs differ. 2. The function is undefined at \(x=1\), and therefore is not continuous on \([0,2]\). 3. Bisection's sign-change guarantee needs continuity throughout the bracket. The expression \(1/(x-1)\) is never zero, so the interval contains no root.

\(f\) is discontinuous at \(x=1\). Opposite endpoint signs do not guarantee a root without continuity; this function has no zero.

- How does the starting interval's width compare with the desired accuracy? - What happens to the uncertainty in the root's location after another bisection? - Would the full bracket width or half its width be the relevant bound for its midpoint?

1. The starting interval has width \(1\). Each bisection step halves its width. 2. After \(n\) steps, the remaining bracket is \(\frac{1}{2^n}\) wide, so its midpoint is at most \(\frac{1}{2^{n+1}}\) from a bracketed root. 3. For \(n=5\), the bound is \(\frac{1}{64}=0.015625\), which is too large. For \(n=6\), it is \(\frac{1}{128}=0.0078125\), which meets the requirement.

\(6\) bisection steps.

- What slope must a parallel tangent have? - How could a tangent-slope requirement be restated as a root-finding problem? - Check which half of each interval still has endpoint values of opposite signs. - Stop when the entire remaining bracket determines the same two-decimal answer.

1. Since the line \(h\) has slope \(3\), the tangent must satisfy \(g'(x)=3\). Differentiate: \(g'(x)=x^4-4x\), so define \(F(x)=g'(x)-3=x^4-4x-3\). 2. The endpoint values are \(F(1.5)=-3.9375<0\) and \(F(2.5)=26.0625>0\). Continuity guarantees a zero inside the bracket. Also, \(F'(x)=4x^3-4>0\) on the interval, so it is the only zero there. 3. Midpoint tests first retain \([1.5,2]\), then \([1.75,2]\), then \([1.75,1.875]\), each with opposite endpoint signs. 4. Continue halving and keeping a sign-changing bracket. Eventually \(1.7841796875<x^*<1.78466796875\). 5. Both endpoints round to \(1.78\) to two decimal places, and so does every value between them. Thus \(x^*\approx1.78\).

\(x^*\approx1.78\).

- Keep only the half-interval whose endpoint function values have opposite signs. - Do not stop merely because the bracket is small; check that every value in it rounds the same way. - Retain enough decimal places in the midpoint calculations.

1. The successive retained brackets are \([2.5,3]\), \([2.5,2.75]\), \([2.625,2.75]\), \([2.625,2.6875]\), \([2.625,2.65625]\), \([2.640625,2.65625]\), \([2.640625,2.6484375]\), \([2.64453125,2.6484375]\), \([2.64453125,2.646484375]\), and \([2.6455078125,2.646484375]\). 2. Every value in the final bracket rounds to \(2.65\) to two decimal places. 3. Therefore, \(\sqrt7\approx2.65\).

Final bracket: \([2.6455078125,2.646484375]\). \(\sqrt7\approx2.65\).

- What do the original endpoint signs establish, and what do they not establish? - Could changing the bracket make the sign test informative? - How could you verify that each proposed subinterval meets the conditions for bisection? - What information do the resulting midpoints provide?

1. The endpoint values are \(f(0)=2\) and \(f(3)=2\). They have the same sign, so the usual sign-change condition on \([0,3]\) fails. 2. At \(x=\frac{3}{2}\), \(f(x)=-\frac{1}{4}\). Thus \([0,\frac{3}{2}]\) and \([\frac{3}{2},3]\) each have opposite endpoint signs. 3. The first midpoints are \(\frac{3}{4}\) and \(\frac{9}{4}\), respectively.

a) \(f(0)=f(3)=2>0\), so the full interval does not have opposite endpoint signs, despite containing roots at \(1\) and \(2\). b) Use \([0,\frac{3}{2}]\) and \([\frac{3}{2},3]\); first midpoints \(\frac{3}{4}\) and \(\frac{9}{4}\).