P3 6.1 - Locating roots

P3 6.1 - Locating roots

Locate solutions of equations by evaluating a function on either side of a root. You will refine a sign-changing interval and learn why a break in a graph can invalidate an apparently convincing argument.

Continuous sign-changing intervals

Can two function values prove that an equation has a solution, even when neither value is zero? For a continuous function, a negative value followed by a positive value forces the graph to cross the axis somewhere between them.

For example, f(x)=x3−x−2f(x)=x^3-x-2 gives f(1)=−2f(1)=-2 and f(2)=4f(2)=4. A polynomial's graph has no breaks: it cannot pass from below the axis to above it without reaching the axis.

Root and continuous bracket

A root of f(x)=0f(x)=0 is an input at which the function equals zero. A function is continuous on an interval when its graph has no gaps, jumps or breaks there, including at its endpoints.

If a<ba<b, ff is continuous on [a,b][a,b], and f(a)f(a) and f(b)f(b) have opposite signs, there is at least one root in (a,b)(a,b). The interval brackets a root.

The strict sign change can be written f(a)f(b)<0f(a)f(b)<0. If an endpoint value is exactly zero, you have found a root at that endpoint instead. A small nonzero value is still a sign, not an exact root.

Worked example 1: a polynomial bracket

Show that x3−x−2=0x^3-x-2=0 has a root between 11 and 22.

Plan. Evaluate the polynomial at both endpoints, then use continuity and the opposite signs.

Let f(x)=x3−x−2f(x)=x^3-x-2. Then

f(1)=1−1−2=−2,f(2)=8−2−2=4.f(1)=1-1-2=-2, \qquad f(2)=8-2-2=4.

Answer. ff is continuous on [1,2][1,2] because it is a polynomial, and f(1)<0<f(2)f(1)<0<f(2). Hence there is a root in (1,2)(1,2).

Check. f(1)f(2)=(−2)(4)=−8<0f(1)f(2)=(-2)(4)=-8<0, confirming opposite signs. The conclusion is existence of a root, not its exact location.

Change only the constant term from −2-2 to −3-3. Recalculate the signs rather than assume the old bracket still works.

Worked example 2: the function changes

Show that x3−x−3=0x^3-x-3=0 has a root between 11 and 22.

Plan. Use the same endpoints for the new continuous polynomial.

Let g(x)=x3−x−3g(x)=x^3-x-3. Then

g(1)=1−1−3=−3,g(2)=8−2−3=3.g(1)=1-1-3=-3, \qquad g(2)=8-2-3=3.

Answer. gg is continuous on [1,2][1,2] and changes sign, so there is at least one root between 11 and 22.

Check. Subtracting 11 from the previous function subtracts 11 from each endpoint value: −2-2 becomes −3-3 and 44 becomes 33.

State what the values prove in a concluding sentence.

Exam tip

Give both function values, identify their opposite signs, state continuity on the interval and conclude that a root lies between the endpoints. A list of calculator outputs leaves the reason unfinished.

Change the constant once more and justify the conclusion.

Refining a bracket

A narrower continuous sign-changing interval gives a more precise location. Evaluate inside the old bracket and retain two endpoints whose function values still have opposite signs.

For f(x)=x3−x−2f(x)=x^3-x-2,

f(1.5)=−0.125,f(1.6)=0.496.f(1.5)=-0.125,\qquad f(1.6)=0.496.

So a root lies between 1.51.5 and 1.61.6. Testing tenths has improved the original bracket (1,2)(1,2); testing hundredths will narrow it further. You may scan values or repeatedly divide an interval into smaller pieces. The justification is the same each time.

Refining a bracket

To refine a bracket, replace it by a smaller interval inside it on which continuity and opposite endpoint signs still guarantee a root. Keep enough digits in function values to establish their signs reliably.

Avoid rounding a small negative value to 00: that would hide the information the method needs.

Worked example 3: hundredths

A root of f(x)=x3−x−2=0f(x)=x^3-x-2=0 lies in (1.5,1.6)(1.5,1.6). Show that a root lies in (1.52,1.53)(1.52,1.53).

Plan. Test the proposed tighter endpoints in the original function.

f(1.52)=1.523−1.52−2=−0.008192f(1.53)=1.533−1.53−2=0.051577\begin{aligned} f(1.52)&=1.52^3-1.52-2=-0.008192\\ f(1.53)&=1.53^3-1.53-2=0.051577 \end{aligned}

Answer. The polynomial remains continuous, and the signs are negative then positive, so a root lies between 1.521.52 and 1.531.53.

Check. The new interval is inside (1.5,1.6)(1.5,1.6), and its width 0.010.01 is one tenth of the old width 0.10.1.

Change only the refinement level: use thousandths in the same function.

Continue this lesson.

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