Intermediate Value Theorem Math Example 3

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Example 3

easy
Show that h(x)=x2−2h(x) = x^2 - 2 has a root in (1,2)(1, 2).

Solution

  1. 1
    hh is continuous (polynomial). h(1)=−1<0h(1) = -1 < 0, h(2)=2>0h(2) = 2 > 0.
  2. 2
    By IVT, ∃ c∈(1,2)\exists\, c \in (1,2) with h(c)=0h(c) = 0. (That root is c=2c = \sqrt{2}.)

Answer

By IVT, 2∈(1,2)\sqrt{2} \in (1, 2).
The IVT gives a clean existence proof for 2\sqrt{2}: since x2−2x^2 - 2 changes sign on (1,2)(1,2), a root must lie there.

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 1 easy

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

Example 2 medium

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