Free math worksheets

- 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.

- 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 \([1.2,1.4]\).

The student 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 \(x\approx1.77\).

- Parallel lines have equal slopes. - Set the first derivative equal to the line's slope. - Use endpoint signs to decide which half-interval still contains the zero. - Continue until the interval determines the answer to two decimal places.

1. Differentiate: \(g'(x)=x^4-4x\). The line \(h\) has slope \(3\), so solve \(F(x)=x^4-4x-3=0\). 2. Since \(F(1.5)=-3.9375<0\) and \(F(2.5)=26.0625>0\), a zero lies in \([1.5, 2.5]\). 3. Repeatedly halve the interval and retain the half whose endpoints have opposite signs. The successive brackets include \([1.5, 2]\), \([1.75, 2]\), and \([1.75, 1.875]\). 4. Continuing the bisection process gives the bracket \([1.7841796875, 1.78466796875]\). Every value in this bracket rounds to \(1.78\) to two decimal places. 5. Therefore, \(x^*\approx1.78\).

\(x^*\approx1.78\)

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