1.09a, 1.09b - Root intervals, accuracy and failed sign tests

1.09a, 1.09b - Root intervals, accuracy and failed sign tests

How can you trust a root that you cannot find exactly? You will locate roots and justify their accuracy under 1.09a, then investigate touching roots, asymptotes and hidden crossings under 1.09b.

What a sign change establishes

Where does x3+x=1x^3+x=1 balance? At x=0x=0 the left side is too small; at x=1x=1 it is too large. The graph between them supplies the missing argument.

You already know

A root of f(x)=0f(x)=0 is an xx-intercept of y=f(x)y=f(x). A strictly increasing function cannot take the same value at two different inputs.

We need a graph that joins up throughout the interval, rather than two disconnected branches.

Root bracket

An interval known to contain a root is a bracket. A sign-change bracket has opposite nonzero endpoint signs and a function continuous throughout the interval.

Continuity means the graph has no break there. It cannot move from below the axis to above it without meeting the axis.

Sign-change test

If ff is continuous on [a,b][a,b] and

f(a)f(b)<0f(a)f(b)<0

then at least one root lies in (a,b)(a,b). This does not establish uniqueness.

Not in the formula sheet: learn it.

Put every term on one side so that zero is the target.

Worked example 1: locating the balance

Question. Show that x3+x=1x^3+x=1 has exactly one real solution in (0,1)(0,1).

Plan. Evaluate the difference at the endpoints, then use strict increase for uniqueness.

f(x)=x3+x−1move the target leftf(0)=−1<0,f(1)=1>0opposite signsf′(x)=3x2+1>0strict increase\begin{aligned} f(x)&=x^3+x-1 && \text{move the target left}\\ f(0)&=-1<0,\qquad f(1)=1>0 && \text{opposite signs}\\ f'(x)&=3x^2+1>0 && \text{strict increase} \end{aligned}

Why two arguments? Continuity and opposite signs establish existence; strict increase rules out a second root.

Answer. The polynomial is continuous, so a root lies in (0,1)(0,1), and that root is unique.

Check. f(0.5)=−0.375f(0.5)=-0.375, consistent with a root farther right.

Change only the target from 11 to 33. The bracket moves with it.

Worked example 2: moving the target

Question. Show that x3+x=3x^3+x=3 has exactly one solution in (1,2)(1,2).

Plan. Use the same difference function with a new constant.

g(x)=x3+x−3g(1)=−1<0,g(2)=7>0opposite signsg′(x)=3x2+1>0same strict increase\begin{aligned} g(x)&=x^3+x-3\\ g(1)&=-1<0,\qquad g(2)=7>0 && \text{opposite signs}\\ g'(x)&=3x^2+1>0 && \text{same strict increase} \end{aligned}

Why change the function? A zero of x3+x−1x^3+x-1 solves the earlier problem, not this one.

Answer. Continuity gives a root in (1,2)(1,2); strict increase makes it unique.

Check. g(1.5)=1.875>0g(1.5)=1.875>0, so (1,1.5)(1,1.5) is a smaller bracket.

The test also works for x+ln⁡x−2x+\ln x-2 on [1,2][1,2]: the endpoint values are −1-1 and ln⁡2>0\ln2>0, and the logarithm is continuous there. Check its positive-input domain first. An exact zero at an endpoint already gives a root; a tiny value rounded to zero does not.

Exam tip

Display the inputs and their signs, then state what follows. A calculator root alone does not show a sign change. Give a separate reason if the question asks for “exactly one”.

Try one further change of target.

Turning a bracket into an accuracy guarantee

A decimal display suggests an answer; a suitable bracket proves which way the root rounds. For the opening cubic, f(0.68)=−0.005568f(0.68)=-0.005568 and f(0.69)=0.018509f(0.69)=0.018509. This bracket crosses 0.6850.685, so it does not yet distinguish rounding to 0.680.68 from rounding to 0.690.69.

Rounding boundaries

For a proposed value rr to dd decimal places, the nearest-rounding boundaries are r−1210−dr-\frac12 10^{-d} and r+1210−dr+\frac12 10^{-d}. Every number strictly between them rounds to rr.

An interval's position matters as well as its width.

Worked example 3: proving two decimal places

Question. Justify the opening root as 0.680.68 to two decimal places.

Plan. Refine the sign bracket until it lies inside (0.675,0.685)(0.675,0.685).

f(0.682)=−0.000785432<0lower boundf(0.683)=0.001611987>0upper bound0.675<0.682<α<0.683<0.685rounding interval\begin{aligned} f(0.682)&=-0.000785432<0 && \text{lower bound}\\ f(0.683)&=0.001611987>0 && \text{upper bound}\\ 0.675&<0.682<\alpha<0.683<0.685 && \text{rounding interval} \end{aligned}

Why these inputs? Both are inside the rounding interval, and opposite signs trap the root between them.

Answer. Continuity gives 0.682<α<0.6830.682<\alpha<0.683, so α=0.68\alpha=0.68 to two decimal places.

Check. Both endpoints round to 0.680.68.

Keep the equation and change only the requested accuracy to three decimal places.

Worked example 4: one more decimal place

Question. Justify α=0.682\alpha=0.682 to three decimal places.

Plan. Test its rounding boundaries.

f(0.6815)=−0.001982606625<0f(0.6825)=0.000412765625>00.6815<α<0.6825continuity and sign change\begin{aligned} f(0.6815)&=-0.001982606625<0\\ f(0.6825)&=0.000412765625>0\\ 0.6815&<\alpha<0.6825 && \text{continuity and sign change} \end{aligned}

Why half a unit? Three decimal places have rounding tolerance 0.00050.0005, not 0.0010.001.

Answer. The root is 0.6820.682 to three decimal places.

Check. The earlier bracket crossed 0.68250.6825, so it could not justify this extra digit.

Absolute error is a different request from decimal places.

Midpoint error bound

If a<α<ba<\alpha<b, the midpoint m=(a+b)/2m=(a+b)/2 satisfies

∣m−α∣<b−a2.|m-\alpha|<\frac{b-a}{2}.

Not in the formula sheet: learn it.

Thus (0.682,0.683)(0.682,0.683) gives midpoint 0.68250.6825 with error less than 0.00050.0005. It does not make every displayed digit correct. Similarly, for x3−2x^3-2, the values at 1.2551.255 and 1.2651.265 are −0.023343625-0.023343625 and 0.0242846250.024284625: its positive root is 1.261.26 to two decimal places.

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