Intermediate Value Theorem Math Example 1

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Example 1

easy
Show that f(x)=x3−x−1f(x) = x^3 - x - 1 has a root in the interval (1,2)(1, 2).

Solution

  1. 1
    ff is a polynomial, hence continuous on [1,2][1, 2].
  2. 2
    f(1)=1−1−1=−1<0f(1) = 1 - 1 - 1 = -1 < 0.
  3. 3
    f(2)=8−2−1=5>0f(2) = 8 - 2 - 1 = 5 > 0.
  4. 4
    Since ff is continuous, f(1)<0<f(2)f(1) < 0 < f(2), by the IVT there exists c∈(1,2)c \in (1,2) with f(c)=0f(c) = 0.

Answer

By IVT, ff has at least one root in (1,2)(1, 2).
The IVT requires continuity and a sign change. Polynomials are continuous everywhere, so verifying the sign change at the endpoints is sufficient.

About Intermediate Value Theorem

If ff is continuous on the closed interval [a,b][a, b] and NN is any value between f(a)f(a) and f(b)f(b), then there exists at least one cc in (a,b)(a, b) such that f(c)=Nf(c) = N.

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More Intermediate Value Theorem Examples

Example 2 medium

Use the IVT to show that [formula] has a solution in [formula].

Example 3 easy

Show that [formula] has a root in [formula].

Example 4 medium

Can IVT be applied to [formula] on [formula] to conclude it attains the value 0? Explain.

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