Calculus Derivatives and Concavity Concepts

Calculus Derivatives and Concavity Concepts

Assessment

Interactive Video

Mathematics

11th Grade - University

Hard

Created by

Liam Anderson

FREE Resource

The video tutorial covers finding maxima and minima in single variable calculus using derivatives and concavity. It then transitions to multivariable calculus, explaining the concept of gradients and flat tangent planes. An example is provided to find where the gradient equals zero, followed by an analysis using second partial derivatives to determine local minima and saddle points. The video concludes by discussing the limitations of the current analysis and hints at the importance of mixed partial derivatives, which will be covered in a subsequent video.

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10 questions

Show all answers

1.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

In single-variable calculus, what is the purpose of setting the derivative of a function to zero?

To identify points with flat tangent lines

To determine the function's rate of change

To calculate the function's integral

To find points where the function is undefined

2.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What does the second derivative test help determine in single-variable calculus?

The function's domain

The concavity of the function

The function's range

The function's asymptotes

3.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

In multivariable calculus, what does finding where the gradient equals the zero vector help identify?

Points of inflection

Horizontal asymptotes

Flat tangent planes

Vertical asymptotes

4.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the main focus of the second partial derivative test in multivariable calculus?

Finding the function's domain

Determining the function's range

Identifying local maxima, minima, or saddle points

Calculating the function's integral

5.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

In the given example, what is the function used to demonstrate finding gradient zero points?

f(x, y) = x^2 + y^2

f(x, y) = x^2y - y^2x

f(x, y) = x^3 - 3xy + y^3

f(x, y) = x^4 - 4x^2 + y^2

6.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What are the solutions for x when solving the partial derivative equations in the example?

x = 0, x = ±√3

x = 0, x = ±1

x = 0, x = ±√2

x = 0, x = ±2

7.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

How is concavity determined for the function in the example using second partial derivatives?

By finding the function's integral

By evaluating the first partial derivatives

By analyzing the function's graph

By calculating the second partial derivatives

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