Locating a root and justifying a bracket — 9709 P3
Use a sign change together with continuity to locate a root.
Notes and key methods
If a continuous f has opposite signs at two inputs, the intermediate value theorem guarantees a root between them. A sign change alone does not establish uniqueness; a monotonicity argument can do that.
Bisection tests the midpoint and retains the half with an opposite-sign endpoint pair. Keep a bracket before reporting rounded digits. A discontinuity can produce a sign change without a root.
Check the conditions
The function is continuous on the full bracket.
A zero endpoint is already a root, not an opposite-sign pair.
Three worked examples
Example 1 · Locating a numerical root
Let f(x)=\(x^{3} + x - 132\) and n=5. Investigate the root between n and n+1.
Prove that there is exactly one root in (n,n+1).
Perform one bisection step and state the new bracket.
Find f′(n).
Find f at the bracket midpoint.
Show the reasoning
(a)
f(n)=-2<0 and f(n+1)=90>0. The polynomial is continuous and its derivative 3x²+1 is positive everywhere.
Answer: \(\left(5, 6\right)\)
(b)
Evaluate the midpoint; retain precisely the half whose endpoints still have opposite signs.
Answer: \(\left(5, \frac{11}{2}\right)\)
(c)
Differentiate the polynomial to help investigate uniqueness.
\(76\)
Answer: \(76\).
(d)
Evaluate at (2n+1)/2 before selecting the next half.
\(\frac{319}{8}\)
Answer: \(\frac{319}{8}\).
Example 2 · Locating a numerical root
Let f(x)=\(x^{3} + x - 5\) and n=1. Investigate the root between n and n+1.
Prove that there is exactly one root in (n,n+1).
Perform one bisection step and state the new bracket.
Find f′(n).
Find f at the bracket midpoint.
Show the reasoning
(a)
f(n)=-3<0 and f(n+1)=5>0. The polynomial is continuous and its derivative 3x²+1 is positive everywhere.
Answer: \(\left(1, 2\right)\)
(b)
Evaluate the midpoint; retain precisely the half whose endpoints still have opposite signs.
Answer: \(\left(\frac{3}{2}, 2\right)\)
(c)
Differentiate the polynomial to help investigate uniqueness.
\(4\)
Answer: \(4\).
(d)
Evaluate at (2n+1)/2 before selecting the next half.
\(- \frac{1}{8}\)
Answer: \(- \frac{1}{8}\).
Example 3 · Locating a numerical root
Let f(x)=\(x^{3} + x - 11\) and n=2. Investigate the root between n and n+1.
Prove that there is exactly one root in (n,n+1).
Perform one bisection step and state the new bracket.
Find f′(n).
Find f at the bracket midpoint.
Show the reasoning
(a)
f(n)=-1<0 and f(n+1)=19>0. The polynomial is continuous and its derivative 3x²+1 is positive everywhere.
Answer: \(\left(2, 3\right)\)
(b)
Evaluate the midpoint; retain precisely the half whose endpoints still have opposite signs.
Answer: \(\left(2, \frac{5}{2}\right)\)
(c)
Differentiate the polynomial to help investigate uniqueness.
\(13\)
Answer: \(13\).
(d)
Evaluate at (2n+1)/2 before selecting the next half.
\(\frac{57}{8}\)
Answer: \(\frac{57}{8}\).
Common mistakes
A sign change across a pole does not guarantee a zero.
A small function value is not by itself a bound on root error.