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) = x3 - 2x - 1
Solution f(x) = x3 - 2x - 1
f(x): Polynomial
→ f(x) is continuous in (1, 2).
f(1) = 13 - 2⋅1 - 1
= 1 - 2 - 1
= -2
→ f(1): (-)
f(2) = 23 - 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).
→ f(x) is continuous in (1, 2).
f(1) = 13 - 2⋅1 - 1
= 1 - 2 - 1
= -2
→ f(1): (-)
f(2) = 23 - 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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