The Intermediate Value Theorem (IVT) applies to continuous functions on a closed interval. It guarantees that the function attains every value between f(a) and f(b).
Statement: If (f) is continuous on [a, b] and (d) is any real number between f(a) and f(b), then there exists at least one c ∈ (a, b) such that f(c) = d.
A continuous function cannot skip values between f(a) and f(b). If f(a) and f(b) have opposite signs, then there exists at least one c ∈ (a,b) such that f(c) = 0.
Proof
The completeness property of real numbers states that every non-empty set of real numbers that has an upper bound also has a smallest upper bound, called the supremum (least upper bound).
We have a continuous function f defined on the interval [a, b], and d lies between f(a) and f(b). To prove that there exists c ∈ [a, b] such that f(c) = d, define a set,
S = {x ∈ [a, b] | f(x) ≤ d}
Now, the upper bound of set S is b, since every x∈[a, b] satisfies x ≤ b. By the completeness property, S has a supremum
c = sup S
Let f(c) ≠ d, then two cases arise,
- If f(c) < d, it implies that there exists an ε > 0 such that for all x ∈ (c , c+ε) ∩ [a , b], f(x) < d.
This contradicts c = sup S as it implies there are points greater than c but still in S. - If f(c) > d, it implies that there exists an ε>0 such that for all x ∈ (c-ε , c) ∩ [a , b], f(x) > d.
This contradicts c = sup S, as there are no points of S arbitrarily close to c from the left.
From the contradictions proved above, therefore f(c) = d.
Hence, we have proved that there exists c ∈ [a, b] such that f(c) = d, which proves the Intermediate Value Theorem.
Applications of Intermediate Value Theorem
Various applications of the Intermediate Value Theorem are as follows:
- Used to check the existence of roots of a continuous function in an interval.
- Helps determine whether a function attains a given value in an interval.
- Applied in numerical methods such as the Bisection Method.
- Forms the basis for other calculus theorems, including the Mean Value Theorem.
- Used to identify critical points where the derivative attains a specific value.
Limitations of Intermediate Value Theorem
These are limitations of IVT, discussed as follows:
- The theorem does not provide any information on number of roots in the interval for the function.
- The theorem is applicable to continuous functions only and is uncertain for discontinuous or piecewise functions.
- The theorem does not give any information about the behaviour of function outside the specified interval.
Solved Examples
Example 1: Check whether the function defined as f(x) = x3 - 8 has a root in the interval [0,4].
Solution:
Here, we have,
- f(0) = 0 - 8 = -8, and
- f(4) = 43 - 8 = 64 - 8 = 56.
As the function yields values with opposite signs at the endpoints of the given interval, by intermediate value theorem, it implies that the function has at least one root in the interval.
Example 2: Show that the function defined as f(x) = ex - 3x has a root in the interval [0,1].
Solution:
To show whether the function has a root in the given interval, we check the value of function at the endpoints of the interval,
We have, f(0) = e0 - 3(0) = 1 - 0 = 1, and
f(1) = e - 3 = -0.28 (approx.)
Thus, we see that function has opposite signed values at endpoints of the interval, so it has at least one root in the interval.
Practice Problems
Q1: Check whether the function defined as f(x) = x2 - 2x has a root in the interval [0, 1].
Q2: Show that the function defined by f(x) = 1 - 2sin(x) has at least one root in the interval [0, π/2].
Q3: Consider the function f(x) = x3 - x + 2, check whether it has a root in the interval [1, 4].
Q4: Check whether the function given by f(x) = 4x - ex has a root in the interval [0, 1].
Q5: Show that the function defined as f(x) = x5 - x has at least one root in the interval [-1, 1].