Understand · explore · practise
Locate roots using continuity and a change of sign. Learn what endpoint values prove, why repeated roots and asymptotes need care, and how to establish uniqueness. Includes interactive graphs and original worked practice.
Before you startFunction values; graphs; domains; basic differentiation
01 / Use two values to trap a root
Continuous on [a, b] and f(a)f(b) < 0 → at least one root in (a, b)
If a continuous graph goes from below the x-axis to above it, or the reverse, it must reach the axis somewhere in between. The crucial conditions are continuity throughout the closed interval and strictly opposite endpoint signs.
A root may cross the axis or only touch it. The sign-change test guarantees existence when its conditions hold; it is not a test that detects every possible root.
f(1) = −1; f(2) = 5.
The function is continuous on [1, 2] and the endpoint values have opposite signs. At least one root lies strictly between 1 and 2.
On [1, 2], f′(x) = 3x² − 1 > 0, so this root is unique within that interval.
The shaded strip is the interval being tested. The gold points are its endpoint values. Curves are shown only inside the plotted window; a dashed purple line marks an excluded input.
02 / Write a complete sign-change argument
Let f(x) = x³ − x − 1
This polynomial is continuous on [1.3, 1.4].
f(1.3) = 2.197 − 1.3 − 1 = −0.103
The left endpoint value is negative.
f(1.4) = 2.744 − 1.4 − 1 = 0.344
The right endpoint value is positive.
There is at least one root α with 1.3 < α < 1.4
Continuity and the change of sign justify the conclusion.
Pause, replay or seek freely. The notes explain the same idea and stay in view.
Show that x² − 5 = 0 has a root in (2, 3).
Evaluate the continuous polynomial at both endpoints.
f(2) = −1 and f(3) = 4. Since f is continuous on [2, 3], the opposite signs guarantee at least one root between 2 and 3.
Show that 2ˣ − 5 = 0 has a root in (2, 3).
2² = 4 and 2³ = 8.
The continuous function has endpoint values −1 and 3. Therefore at least one root lies in (2, 3).
03 / Check for a gap before using the signs
f(1) = −1 and f(3) = 1
The endpoint signs are opposite.
x = 2 is excluded from the domain
The function is not continuous on the whole interval.
1/(x − 2) can never equal zero
The nonzero numerator rules out every root.
The sign-change theorem cannot be applied here
The graph switches sides through an asymptote, not through a root.
A discontinuity removes the guarantee; it does not automatically prove that the interval has no root. Investigate the actual function.
For g(x) = 1/x + 2, explain why the sign-change theorem cannot be applied on [−1, 1], and find any root in that interval.
x = 0 is excluded. Solve 1/x + 2 = 0 separately.
Continuity fails at zero, so the theorem is unavailable on [−1, 1]. Nevertheless g has the root x = −1/2 in the interval. A discontinuity does not prove the absence of roots.
Let f(x) = −1 for x < 0 and f(x) = 1 for x ≥ 0. Its values at −1 and 1 have opposite signs. Does it have a root?
List the only outputs this function can have.
No. Its outputs are only −1 and 1. It is discontinuous at zero, so the sign-change theorem does not apply.
04 / Equal endpoint signs do not rule out roots
f(x) = (x − 1)² has f(0) = f(2) = 1
The endpoint signs agree.
Yet f(1) = 0
The repeated root touches the axis without crossing.
g(x) = (x − 1)² + 1 also has positive endpoint values
But g(x) ≥ 1, so it has no root.
Endpoint signs alone do not distinguish these situations
Use other information, such as factorisation, a graph or a bound.
For f(x) = x² − 1, evaluate f(−2) and f(2), then find the roots in (−2, 2).
Factor x² − 1.
Both endpoint values are 3. The roots are x = −1 and x = 1. Two crossings can return the graph to the same side of the axis.
Explain why x² + 4 has no real root.
Use the fact that x² ≥ 0.
x² + 4 ≥ 4 for every real x, so it cannot be zero. This is a global lower-bound argument, not an inference from two endpoint signs.
05 / A sign change may contain several roots
f(−1.5) = −1.875 and f(1.5) = 1.875
Continuity and opposite signs guarantee a root.
f(x) = x(x − 1)(x + 1)
Factorisation gives roots −1, 0 and 1.
All three lie within the interval
One sign bracket does not prove uniqueness.
Smaller disjoint sign brackets can certify several distinct roots
Use a separate interval around each crossing.
For x³ − x, calculate the values at −1.5, −0.5, 0.5 and 1.5. What do the three adjacent intervals prove?
The signs alternate.
The values are −1.875, 0.375, −0.375 and 1.875. Continuity gives a root in each of (−1.5, −0.5), (−0.5, 0.5) and (0.5, 1.5). These open intervals are disjoint, so the roots are distinct.
Two overlapping sign brackets each contain a root. Does this prove there are two distinct roots?
The same root may belong to both intervals.
No. For example, x − 1 changes sign on both [0, 2] and [0.5, 1.5], but has only one root. Disjoint brackets avoid this particular overlap problem.
06 / Deal with a zero endpoint directly
f(2) = 0
x = 2 is an exact root at the endpoint.
f(3) = 5
The product of endpoint values is zero, not negative.
This does not guarantee a root strictly inside (2, 3)
In fact x² − 4 = (x − 2)(x + 2) has none there.
Distinguish the closed interval [2, 3] from the open interval (2, 3)
An endpoint belongs only to the closed interval.
If f(a) = 0 and f(b) > 0, with a < b, what root is known immediately?
Use the definition of a root.
x = a is a root in [a, b]. These two endpoint values alone do not establish any root in (a, b).
07 / Add a separate argument for uniqueness
f(1) = −2 and f(2) = 6
Continuity gives at least one root in (1, 2).
f′(x) = 3x² + 1 > 0 for every real x
The function is strictly increasing.
A strictly increasing function can take the value zero at most once
Two zeros would contradict strict increase.
There is exactly one real root, and it lies in (1, 2)
Existence and uniqueness have both been justified.
Explain why the root of x³ − x − 1 in (1.3, 1.4) is unique within that interval.
Study f′ on [1.3, 1.4].
f′ = 3x² − 1 > 0 there, so f is strictly increasing. Together with the known sign bracket, this proves exactly one root in that interval.
Show that e^(−x) − x/2 has exactly one root in (0, 2).
Check endpoints and the derivative.
f(0) = 1 > 0 and f(2) = e⁻² − 1 < 0. The function is continuous and f′ = −e^(−x) − 1/2 < 0. Therefore there is exactly one root in (0, 2).
08 / Turn an intersection into a root problem
Let f(x) = ln x + x − 3
An intersection corresponds exactly to f(x) = 0.
f(1) = −2 and f(3) = ln 3 > 0
The function is continuous on [1, 3].
There is a root in (1, 3)
Equivalently, the two graphs intersect there.
f′(x) = 1/x + 1 > 0 for x > 0
This intersection is the only one on the positive domain.
Show that y = √x and y = 3 − x intersect at least once for 1 < x < 3.
Use f(x) = √x + x − 3.
f is continuous on [1, 3], f(1) = −1 and f(3) = √3 > 0. Therefore a root, and hence an intersection, lies in (1, 3).
09 / A derivative can be the function whose root you locate
y′ = x² − 2
Stationary inputs solve y′ = 0, not y = 0.
y′(1) = −1 and y′(2) = 2
The derivative is continuous on [1, 2].
There is a stationary input in (1, 2)
The sign-change theorem is applied to y′.
Here y″ = 2x > 0 on the interval
The derivative increases through zero, so the stationary point is a minimum.
For y = x³/3 − 3x, show that a stationary input lies between 1 and 2, and classify that point.
Use y′ = x² − 3 and y″ = 2x.
y′(1) = −2 and y′(2) = 1. Since y′ is continuous, a zero lies between them. Also y″ > 0 there, so the stationary point is a minimum.
10 / Say precisely what your evidence establishes
A student finds f(2) < 0 and f(3) > 0 and writes “There is exactly one root, equal to 2.5.” What is missing?
Consider continuity, uniqueness and the midpoint.
Continuity on [2, 3] is needed for the sign-change guarantee. Even then it proves at least one root in (2, 3), not uniqueness and not an exact midpoint root. Those require additional evidence.
Section 1 of 10 · Use two values to trap a root