# Intermediate Value Theorem

See how to use the intermediate value theorem.

1 example and its solution.

## Intermediate Value Theorem

### Theorem

then x = c exists that satisfies f(c) = k.

(k is between f(a) and f(b).)

### Example

Show that the zero of f(x) exists in (1, 2).

f(x) = x

Solution f(x) = x

^{3}- 2x - 1 f(x): Polynomial

→ f(x) is continuous in (1, 2).

f(1) = 1

= 1 - 2 - 1

= -2

→ f(1): (-)

f(2) = 2

= 8 - 4 - 1

= 3

→ f(2): (+)

f(1) ≠ f(2)

Then, by the intermediate value theorem,

x = c exists that satisfies f(c) = 0.

(0 is between f(1) and f(2).)

∴ The zero of y = f(x) exists in the interval (1, 2).

→ f(x) is continuous in (1, 2).

f(1) = 1

^{3}- 2⋅1 - 1= 1 - 2 - 1

= -2

→ f(1): (-)

f(2) = 2

^{3}- 2⋅2 - 1= 8 - 4 - 1

= 3

→ f(2): (+)

f(1) ≠ f(2)

Then, by the intermediate value theorem,

x = c exists that satisfies f(c) = 0.

(0 is between f(1) and f(2).)

∴ The zero of y = f(x) exists in the interval (1, 2).

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