Me1, Me2 - Root intervals, accuracy and failed sign tests

Me1, Me2 - Root intervals, accuracy and failed sign tests

Can a table of function values miss two roots between consecutive integers? You will locate roots and choose useful starting intervals under Me1, then explain touching roots, discontinuities and missed crossings under Me2. You will also distinguish a plausible decimal from a justified approximation.

What a sign change establishes

Suppose a calculator table samples h(x)=(x−2.2)(x−2.8)h(x)=(x-2.2)(x-2.8) only at integers. At 22 and 33, it reports 0.160.16 both times. There are nevertheless two roots between them! Before diagnosing that puzzle, establish what a sign test actually promises.

You already know

A root satisfies f(x)=0f(x)=0. Move all terms to one side before testing signs. Strict increase can establish uniqueness, but existence needs its own argument.

The test uses the behaviour throughout the interval, not merely two calculator outputs.

Sign-change bracket

If a function is continuous throughout [a,b][a,b] and has opposite nonzero signs at the endpoints, (a,b)(a,b) contains at least one root. “Continuous” means there is no break in the graph there.

A graph that joins up cannot travel from below the axis to above it without meeting it.

Locating a root

For continuous ff on [a,b][a,b],

f(a)f(b)<0⟹at least one root in (a,b).f(a)f(b)<0\quad\Longrightarrow\quad \text{at least one root in }(a,b).

Not in the formula sheet: learn it.

Such an interval will also help choose a starting value for iteration.

Worked example 1: finding a starting interval

Question. Show that x3−x−4=0x^3-x-4=0 has exactly one root in (1,2)(1,2).

Plan. Evaluate both endpoints, then inspect the derivative on this interval.

f(1)=1−1−4=−4<0f(2)=8−2−4=2>0f′(x)=3x2−1>0for 1≤x≤2\begin{aligned} f(1)&=1-1-4=-4<0\\ f(2)&=8-2-4=2>0\\ f'(x)&=3x^2-1>0 && \text{for }1\le x\le2 \end{aligned}

Why specify the interval for the derivative? It is positive here, but not for every real input.

Answer. The polynomial is continuous, so there is a root in (1,2)(1,2); strict increase makes it the only root there.

Check. f(1.5)=−2.125<0f(1.5)=-2.125<0, so (1.5,2)(1.5,2) also brackets it.

Change only the constant term from −4-4 to −5-5.

Worked example 2: the same interval, a different target

Question. Show that x3−x−5=0x^3-x-5=0 has exactly one root in (1,2)(1,2).

Plan. Repeat the endpoint and monotonicity arguments for the new function.

g(1)=−5<0,g(2)=1>0g′(x)=3x2−1>0on [1,2]\begin{aligned} g(1)&=-5<0,\qquad g(2)=1>0\\ g'(x)&=3x^2-1>0 && \text{on }[1,2] \end{aligned}

Why evaluate afresh? Changing the constant moves the root, even though the derivative stays unchanged.

Answer. Continuity and opposite signs establish a root; strict increase establishes uniqueness in (1,2)(1,2).

Check. g(1.5)=−3.125g(1.5)=-3.125, lower than the corresponding value in example 1.

Logarithms need a domain check too. For x+ln⁡x−2x+\ln x-2, the interval [1,2][1,2] is inside the positive-input domain; the values −1-1 and ln⁡2>0\ln2>0 establish a root. Likewise, x3+x−1x^3+x-1 has signs −1-1 and 11 at 00 and 11.

Exam tip

For “Show that”, identify the function, show the endpoint signs and explain the conclusion. For “exactly one”, add a justified uniqueness argument. An exact endpoint zero gives a root immediately.

Use the same reasoning with one more change of constant.

Turning a bracket into an accuracy guarantee

An interval can locate a root without fixing its rounded value. For f(x)=x3−x−4f(x)=x^3-x-4, the root lies near 1.79631.7963. The whole bracket must fit the accuracy being claimed.

Rounding interval

A value rr to dd decimal places represents numbers between r−1210−dr-\frac12 10^{-d} and r+1210−dr+\frac12 10^{-d}, with exact ties treated by the stated rounding convention. A strictly interior bracket avoids ties.

Keep full precision for function evaluations; round only the values you report.

Worked example 3: justifying a decimal

Question. Justify the root as 1.801.80 to two decimal places.

Plan. Find opposite signs within its rounding interval (1.795,1.805)(1.795,1.805).

f(1.796)=−0.002793664<0f(1.797)=0.005888573>01.795<1.796<α<1.797<1.805\begin{aligned} f(1.796)&=-0.002793664<0\\ f(1.797)&=0.005888573>0\\ 1.795&<1.796<\alpha<1.797<1.805 \end{aligned}

Why retain the final zero? 1.801.80 explicitly gives two decimal places.

Answer. Continuity places the root inside the rounding interval, so α=1.80\alpha=1.80 to two decimal places.

Check. Both 1.7961.796 and 1.7971.797 round to 1.801.80.

Change only the required precision to four decimal places.

Worked example 4: testing rounding boundaries

Question. Justify α=1.7963\alpha=1.7963 to four decimal places.

Plan. Test 1.796251.79625 and 1.796351.79635 in the same function.

f(1.79625)=−0.000624115234375<0f(1.79635)≈0.000243892873>01.79625<α<1.79635continuous sign bracket\begin{aligned} f(1.79625)&=-0.000624115234375<0\\ f(1.79635)&\approx0.000243892873>0\\ 1.79625&<\alpha<1.79635 && \text{continuous sign bracket} \end{aligned}

Why not merely quote two agreeing iterates? Agreement can come from rounding; the bracket establishes the required root location.

Answer. The root is 1.79631.7963 to four decimal places.

Check. The earlier bracket (1.796,1.797)(1.796,1.797) was too wide for this conclusion.

If the task instead asks for an error bound, use the interval width.

Midpoint estimate

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

∣m−α∣<b−a2.|m-\alpha|<\frac{b-a}{2}.

Not in the formula sheet: learn it.

For example, (1.796,1.797)(1.796,1.797) gives midpoint 1.79651.7965 with absolute error less than 0.00050.0005. For the different function x3+x−1x^3+x-1, the bracket (0.682,0.683)(0.682,0.683), with endpoint values −0.000785432-0.000785432 and 0.0016119870.001611987, proves 0.680.68 to two decimal places. The method is unchanged.

Continue this lesson.

You’re previewing NovaLearn without an account. Create a free account to read the rest.

Create free account