Intermediate Value Theorem Math Example 4

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Example 4

medium

Can IVT be applied to

f(x) = \frac{1}{x}

[-1, 1]

to conclude it attains the value 0? Explain.

Solution

1
$f(-1) = -1$ and $f(1) = 1$ , so $0$ is between them.
2
However, $f(x) = 1/x$ is not continuous on $[-1, 1]$ — it has an infinite discontinuity at $x = 0$ .
3
The hypotheses of IVT fail, so the theorem does not apply.
4
Indeed, $f(x) = 1/x \neq 0$ for all $x$ in its domain, confirming IVT cannot be used here.

Answer

No. IVT does not apply because

f

is discontinuous on

[-1, 1]

IVT requires continuity on the entire closed interval. A single discontinuity invalidates the theorem, even if the sign change condition is satisfied at the endpoints.

About Intermediate Value Theorem

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

Learn more about Intermediate Value Theorem →

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 3 easy

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

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Related Concepts

Limit Types of Continuity and Discontinuity Mean Value Theorem