I1 - Locating roots and sign-change limits

I1 - Locating roots and sign-change limits

Locate solutions by finding intervals that contain roots, justify a requested accuracy, and recognise when endpoint signs cannot support a conclusion. A calculator can suggest a decimal; a short argument can establish which digits you may trust.

Continuous root brackets

Where do the curves y=x3y=x^3 and y=x+2y=x+2 meet, and how could you justify the answer without an exact formula? At x=1x=1, the cubic is lower; at x=2x=2, it is higher. Since both curves are unbroken between these inputs, they must meet somewhere in between.

You already know

A root of f(x)=0f(x)=0 is an input at which y=f(x)y=f(x) meets the horizontal axis. To find an intersection g(x)=h(x)g(x)=h(x), form f(x)=g(x)−h(x)f(x)=g(x)-h(x), so that the target becomes zero.

Subtracting the two outputs measures their signed vertical separation. Use the same subtraction at every trial input: switching its order can manufacture a false sign change.

Bracket for a root

A bracket is an interval known to contain a root. If ff is continuous throughout [a,b][a,b] and f(a)f(b)<0f(a)f(b)<0, at least one root lies in (a,b)(a,b). For example, a polynomial with values of opposite signs at two inputs has a root between them. Opposite signs across a break do not give this guarantee.

Here continuous means that there is no break anywhere on the interval. An unbroken curve cannot move from below the axis to above it without passing through zero. This sign-change result is not in the formula booklet: learn it. If an endpoint value equals zero, that endpoint is already a root; the strict sign-change test is unnecessary.

Worked example 1: bracketing an intersection

Question. Use a change of sign to locate a solution of x3=x+2x^3=x+2 between consecutive integers, then narrow the bracket to width 0.010.01.

Plan. Subtract the right side, test the endpoints, then try closer inputs while retaining opposite signs.

f(x)=x3−x−2target is zerof(1)=−2below the axisf(2)=4above the axisf(1.5)=−0.125retain the upper halff(1.52)=−0.008192new lower endpointf(1.53)=0.051577new upper endpoint\begin{aligned} f(x)&=x^3-x-2 && \text{target is zero} \\ f(1)&=-2 && \text{below the axis} \\ f(2)&=4 && \text{above the axis} \\ f(1.5)&=-0.125 && \text{retain the upper half} \\ f(1.52)&=-0.008192 && \text{new lower endpoint} \\ f(1.53)&=0.051577 && \text{new upper endpoint} \end{aligned}

Why this step? Each retained interval has opposite signs at its ends, so continuity keeps the existence argument valid as the bracket narrows.

Answer. The polynomial is continuous and changes sign, so at least one root α\alpha satisfies 1.52<α<1.531.52<\alpha<1.53.

Check. The final width is 1.53−1.52=0.011.53-1.52=0.01; neither endpoint value is zero.

Now change only the constant term, from −2-2 to −3-3. The bracket moves, but the evidence needed for existence stays the same.

Worked example 2: a changed constant

Question. Show, using a change of sign, that x3−x−3=0x^3-x-3=0 has a root in (1.67,1.68)(1.67,1.68).

Plan. Evaluate the same zero-function at both supplied endpoints and state why the signs imply a root.

f(1.67)=1.673−1.67−3substitute the lower input=−0.012537negativef(1.68)=1.683−1.68−3substitute the upper input=0.061632positive\begin{aligned} f(1.67)&=1.67^3-1.67-3 && \text{substitute the lower input} \\ &=-0.012537 && \text{negative} \\ f(1.68)&=1.68^3-1.68-3 && \text{substitute the upper input} \\ &=0.061632 && \text{positive} \end{aligned}

Why this step? The question asks for a sign-change argument, so a calculator's decimal root alone would not establish the specified method.

Answer. The polynomial is continuous throughout [1.67,1.68][1.67,1.68] and its endpoint values have opposite signs. Hence a root lies strictly between 1.671.67 and 1.681.68.

Check. The bracket is consistent with the coarser values f(1)=−3f(1)=-3 and f(2)=3f(2)=3.

Existence means at least one root. A sign change does not count the roots or establish uniqueness; monotonicity can provide that extra information when needed.

Watch out

Wrong: “g(a)<h(a)g(a)<h(a) and h(b)<g(b)h(b)<g(b), so compare g(a)−h(a)g(a)-h(a) with h(b)−g(b)h(b)-g(b).” The subtraction changes order. Use f=g−hf=g-h at both ends, then state opposite signs and continuity on the whole interval.

For your turn, change only the constant again. The supplied interval saves searching time; you still need to justify why it contains a root.

Accuracy from a bracket

A narrow bracket limits error, but the position of its ends relative to rounding boundaries determines the last reported digit.

For the opening cubic, 1.52<α<1.531.52<\alpha<1.53 does not yet prove 1.521.52 to two decimal places: the interval crosses 1.5251.525, where the rounded value changes. Test the half-unit boundaries for the accuracy requested.

Error from a midpoint bracket

If a<α<ba<\alpha<b and m=(a+b)/2m=(a+b)/2, then

∣α−m∣<b−a2|\alpha-m|<\frac{b-a}{2}

Repeatedly testing the midpoint and retaining a continuous sign-changing half is bisection. After kk halvings, the width is (b−a)/2k(b-a)/2^k.
Not in the formula booklet: learn it.

This also produces a useful surprise: ten halvings reduce a bracket's width by a factor of 10241024. From a width of 11, the midpoint error is then less than 1/20481/2048, about 0.00048830.0004883. This controls absolute error; a decimal-place claim still needs a rounding-boundary check.

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