IVT, EVT, MVT

IVT, EVT, MVT

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Flashcard

Mathematics

12th Grade

Hard

Created by

Wayground Content

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15 questions

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1.

FLASHCARD QUESTION

Front

What is the Intermediate Value Theorem (IVT)?

Back

The Intermediate Value Theorem states that if a function is continuous on a closed interval [a, b], then for any value L between f(a) and f(b), there exists at least one c in (a, b) such that f(c) = L.

2.

FLASHCARD QUESTION

Front

What is the Extreme Value Theorem (EVT)?

Back

The Extreme Value Theorem states that if a function is continuous on a closed interval [a, b], then it attains both a maximum and a minimum value at some points in that interval.

3.

FLASHCARD QUESTION

Front

What is the Mean Value Theorem (MVT)?

Back

The Mean Value Theorem states that if a function is continuous on a closed interval [a, b] and differentiable on the open interval (a, b), then there exists at least one c in (a, b) such that f'(c) = (f(b) - f(a)) / (b - a).

4.

FLASHCARD QUESTION

Front

How do you determine if a function has a relative maximum or minimum at a point?

Back

To determine if a function has a relative maximum or minimum at a point, use the first derivative test: if f'(x) changes from positive to negative at x, then f has a relative maximum; if it changes from negative to positive, then f has a relative minimum.

5.

FLASHCARD QUESTION

Front

What is a continuous function?

Back

A continuous function is a function that does not have any breaks, jumps, or holes in its graph. Formally, a function f is continuous at a point x = c if lim (x -> c) f(x) = f(c).

6.

FLASHCARD QUESTION

Front

What is a differentiable function?

Back

A differentiable function is a function that has a derivative at each point in its domain. This means that the function is smooth and has no sharp corners or cusps.

7.

FLASHCARD QUESTION

Front

Provide an example of a function that satisfies the conditions of the Mean Value Theorem on the interval [1, 3].

Back

Consider the function f(x) = x^2. It is continuous on [1, 3] and differentiable on (1, 3). By MVT, there exists c in (1, 3) such that f'(c) = (f(3) - f(1)) / (3 - 1).

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