Locate and bisect a circular-equation root

Before choosing a bisection half, which setup is required?

Function

\(\displaystyle f(x)=(x-a)^2-b^2\)

Interval

\(\displaystyle [a-2b,a+2b]\)

Test the claim that positive endpoint values prevent interior roots. Inspect the function over the interval shown below. Change a or b while keeping b positive. This table gives exact theoretical values

Why is every displayed setting a counterexample to the claim?

Bisection for a circular equation

Form a continuous difference function and check the endpoints for an exact root. Otherwise locate opposite endpoint signs and test the midpoint. Stop if its value is zero; if not, keep the sign-changing half and repeat

1. 1Form a difference function Write the circular side minus the algebraic side so that an intersection is a root of F. \(\displaystyle F(x)=f(x)-P(x),\quad F(x)=0\)
2. 2Check the endpoint signs Check that F is continuous throughout the interval. If an endpoint value is zero, that endpoint is an exact root; otherwise use non-zero endpoint values with opposite signs \(\displaystyle F(L)<0<F(R)\quad\text{or}\quad F(R)<0<F(L)\)
3. 3Calculate the midpoint Average the current left and right endpoints to obtain the next test value \(\displaystyle M=\frac{L+R}{2}\)
4. 4Keep the sign-changing half If F(M) = 0, stop: M is an exact root. Otherwise retain the half whose endpoints have opposite signs and repeat \(\displaystyle F(L)F(M)<0\quad\Rightarrow\quad [L,M]\)

Given

\(\displaystyle M_1=\frac{0+\frac{\pi}{2}}{2}=?\pi\)

Calculate the first midpoint and enter its coefficient of π

Function

\(\displaystyle F(x)=\sin x-\frac{1}{2}\)

Starting interval

\(\displaystyle 0\le x\le\frac{\pi}{2}\)

Given

\(\displaystyle \frac{\pi}{4}-0=?\pi\)

Find the width of the sign-changing half retained after the first midpoint

Keeping the same-sign half. One of these lines is right. Which one?

What a sign change proves

Opposite nonzero endpoint signs guarantee at least one root only when the difference function is continuous throughout the interval. They do not prove uniqueness. If an endpoint is already zero, record that exact root before starting the sign-changing procedure

1. 1Continuity matters The function 1/x has opposite signs at −1 and 1 but no root; its discontinuity at zero prevents the sign-change theorem from applying
2. 2A separate root-count argument Strict monotonicity together with a sign change proves exactly one root. A finite table of signs alone can establish a bracket, but not rule out extra roots

Worked model: cos x = x

Use radians and F(x) = cos x − x on [0,1]. F(0) = 1 and F(1) is about −0.4597. F is continuous and strictly decreasing here: cos x decreases while −x decreases. Therefore this interval has exactly one root

1. 1First midpoint M = 0.5; F(M) is about 0.3776, so keep [0.5,1]
2. 2Second midpoint M = 0.75; F(M) is about −0.0183, so keep [0.5,0.75]
3. 3Third midpoint M = 0.625; F(M) is about 0.1860, so keep [0.625,0.75]. These displayed approximations explain signs; carry unrounded values when running the algorithm

Given

\(\displaystyle L=5/8,\quad R=3/4,\quad M=(L+R)/2\)

Calculate the midpoint of the interval remaining after those three steps

Given

\(\displaystyle L=5/8,\quad R=3/4\)

Give the largest possible distance between that midpoint estimate and a root in the interval

Certify the reported precision

Continuing gives F(0.7390) > 0 and F(0.7391) < 0. Both endpoints round to 0.739 to three decimal places, so every number in the bracket does too. A small interval alone is insufficient if it crosses a rounding boundary

1. 1Keep the contract of the question A question asking for the midpoint after a fixed number of steps wants that estimate. A question asking for a correctly rounded root needs enough accuracy to justify the rounding

Worked model: refute a root claim

Claim: a continuous function with positive values at both ends of an interval has no roots inside. A counterexample must meet the stated conditions and contradict the conclusion. Use the function and interval below

1. 1Plan the technology record Enter (x-0.5)^2-1 with explicit brackets. Use x from −1.5 to 2.5 and y from −1.5 to 3.5. Request a table at −1.5, −0.5, 0.5, 1.5 and 2.5. These settings reveal the endpoints, vertex and crossings
2. 2Expected output to check The corresponding exact values are 3, 0, −1, 0, 3. This is an authored prediction, not a claim that a particular device was run
3. 3Model explanation The polynomial is continuous, and both endpoint values are 3 > 0. Yet its roots are −1/2 and 3/2, both strictly inside the interval. Therefore the claim is false. The graph helps find the example; exact substitution verifies it

Your task: produce and verify a counterexample

Choose your own a from −1, 0, 1, 2 and b from 0.5, 1, 1.5. Use your calculator, spreadsheet or graphing tool to produce a graph and five-row table for f(x) = (x − a)² − b² on [a − 2b,a + 2b]. Choose a vertical window that shows −b² and 3b². Save three settings, changing only a or only b at a time. Compare their graphs and tables and explain what changes and what stays true. Keep your actual input and output; do not substitute the authored table for your technology record

1. 1Written explanation State the claim, your parameter choices, continuity, both endpoint signs and the two interior roots. Explain why one valid counterexample refutes a universal claim. A graph picture alone is not that explanation
2. 2Independent checks Substitute the endpoints and roots by hand. Check that b is positive, the roots are strictly inside and every table value matches your entered rule. If the output differs, investigate brackets or settings before concluding
3. 3Ungraded review Compare your saved record and explanation with the worked model. Ask a teacher or peer to check the conditions and logic. Execution and written reasoning are ungraded; selecting or completing this activity does not establish proficiency

Given

\(\displaystyle f(x)=(x+1)^2-9/4,\quad x=-4\)

Separate check: for a = −1 and b = 3/2, calculate f at the left endpoint a − 2b

Given

\(\displaystyle a=-1,\quad b=3/2\)

For this same fixed example, calculate the smaller root a − b.

Worked model: preserve meaning across notation

The formal expression is shown below. A common linear input is (x^2-16)/(x-4). The ^ denotes a power and the brackets group the entire numerator and denominator. Your device may require different syntax; record its actual input and output

1. 1Simplified output and original domain Factoring gives (x − 4)(x + 4)/(x − 4) = x + 4 only for x ≠ 4. A simplifier may display x + 4 without the exclusion. The original expression remains undefined at 4; the displayed polynomial alone would give 8 there
2. 2An output convention worth checking For real x, sqrt(x^2) is |x|, often written abs(x), not always x. At x = −3 the value is 3. With an assumption x ≥ 0 a tool can legitimately simplify it to x; record assumptions as well as output
3. 3Model comparison At x = 2 both admissible forms of r give 6, but agreement at one input does not prove identity. Factorisation proves agreement on the original domain. Domain restrictions and real-variable assumptions are part of the statement

Your task: translate and compare technology output

Choose a = 2, 3 or 5 and enter (x² − a²)/(x − a) into technology using its supported syntax. If it has symbolic simplification, record that output; otherwise compare a table of the original and candidate x + a at admissible inputs and at x = a. Also inspect sqrt(x²) at a negative and positive input

1. 1Make a three-column record Write the formal expression and domain, your exact technology input/settings, and the actual output. Explain each power, bracket, absolute-value notation and any undefined/error result. Record real-variable assumptions or restrictions you supplied
2. 2Model for the reasoning The two rational forms agree when x ≠ a because the common factor can be cancelled there. The polynomial extends to a value at x = a, while the original does not. For sqrt(x²), a negative test distinguishes |x| from x; factorisation or a sign argument completes the explanation
3. 3Ungraded comparison Your actual device notation, execution and written explanation are ungraded. Check the record with a teacher or peer. The following fixed arithmetic checks give feedback on components, not credit for this external task

Given

\(\displaystyle r(x)=\frac{x^2-25}{x-5},\quad x=-2\)

Separate check: evaluate the original rational expression at x = −2.

For r(x) = (x² − 25)/(x − 5), a tool displays x + 5. What must you preserve?

Given

\(\displaystyle \sqrt{x^2},\quad x=-7/2\)

Evaluate the principal square root at this negative input

Worked model: choose functionality for a purpose

Task: find the unique solution of cos x = x in [0,1] to three decimal places and justify it. Use radians. Numerical, graphical, symbolic and statistical tools answer different questions

1. 1Graphical plan Plot y = cos x and y = x with x in [0,1] and y in [0,1.1]. This helps locate their intersection, but pixel position does not certify three decimal places or uniqueness
2. 2Numerical plan and check Use a root solver on cos(x)−x with bracket [0,1], or implement bisection. Retain more digits than you report. Independently check opposite signs at 0.7390 and 0.7391 and the decreasing difference function, then report 0.739
3. 3Model rationale I chose a numerical solver for a decimal root and a graph to locate the interval. I did not rely on a symbolic exact expression or a plot for accuracy. Statistical sampling has no role in this deterministic equation. I kept the interval and radian setting and verified the solver output with an independent bracket

Your task: choose and justify a solving method

Find the root of the equation below on the stated interval to two decimal places. Choose and justify suitable technology functionality before running it. Record the radian setting, commands, viewing window or initial bracket, unrounded output and reported value

1. 1Produce evidence Graph the two sides or their difference to locate the root. Use numerical solving or your own bisection implementation. Obtain a fresh sign-changing bracket whose endpoints round to the same two-decimal value. Explain why a graphical intersection alone would not certify it
2. 2Independent mathematical check The difference sin x − x/2 is continuous. It is strictly decreasing on [1.5,2], because its derivative cos x − 1/2 is negative throughout. Verify endpoint signs and your final bracket by direct substitution, separately from the root command
3. 3Explain the choice and limitations State what graphical, numerical, symbolic or statistical functionality could contribute, and why your chosen combination meets this task. Your run record, chosen method and written rationale are ungraded; review them against the worked model with a teacher or peer

Given

\(\displaystyle L=1.8954,\quad R=1.8956\)

Separate check: calculate the midpoint of this supplied final bracket

Direct evaluations give F(1.8954) > 0 and F(1.8956) < 0 for F(x) = sin x − x/2. Which report is certified by this bracket?

Review the record, not just the answer

For each technology task, keep the question, assumptions, actual input/settings, output, independent check and explanation. A correct intermediate number or a completed screen does not establish that you independently carried out or justified the process

1. 1What the lesson supplies Worked models, original student tasks, parameter bounds, explicit review criteria and feedback on selected mathematical components
2. 2What remains ungraded Your technology execution, choice of example or functionality, written reasoning and independent performance. Retain that work for discussion; these activities do not add proficiency evidence