IVT, EVT, MVT

IVT, EVT, MVT

Assessment

Flashcard

Mathematics

12th Grade

Hard

Created by

Wayground Content

FREE Resource

Student preview

quiz-placeholder

15 questions

Show all answers

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).

Create a free account and access millions of resources

Create resources

Host any resource

Get auto-graded reports

Google

Continue with Google

Email

Continue with Email

Classlink

Continue with Classlink

Clever

Continue with Clever

or continue with

Microsoft

Microsoft

Apple

Apple

Others

Others

By signing up, you agree to our Terms of Service & Privacy Policy

Already have an account?