zuai-logo
zuai-logo
  1. AP Calculus
FlashcardFlashcard
Study GuideStudy GuideQuestion BankQuestion BankGlossaryGlossary

How to find local extrema using the Second Derivative Test?

  1. Find critical points. 2. Compute the second derivative. 3. Evaluate the second derivative at each critical point. 4. Determine if each point is a local max, min, or neither.
Flip to see [answer/question]
Flip to see [answer/question]
Revise later
SpaceTo flip
If confident

All Flashcards

How to find local extrema using the Second Derivative Test?

  1. Find critical points. 2. Compute the second derivative. 3. Evaluate the second derivative at each critical point. 4. Determine if each point is a local max, min, or neither.

Steps to determine concavity of a function.

  1. Find the second derivative, f′′(x)f''(x)f′′(x). 2. Set f′′(x)=0f''(x) = 0f′′(x)=0 and solve for x. 3. Create a sign chart for f′′(x)f''(x)f′′(x). 4. Determine intervals of concave up (f′′(x)>0f''(x) > 0f′′(x)>0) and concave down (f′′(x)<0f''(x) < 0f′′(x)<0).

How to find inflection points?

  1. Find the second derivative, f′′(x)f''(x)f′′(x). 2. Set f′′(x)=0f''(x) = 0f′′(x)=0 and solve for x. 3. Check if the concavity changes at each potential inflection point.

What to do if the Second Derivative Test is inconclusive?

Use the First Derivative Test or analyze the behavior of the function around the critical point.

How to use the second derivative to determine if x=cx = cx=c is a local max?

Find f′(x)f'(x)f′(x), set f′(c)=0f'(c) = 0f′(c)=0 to confirm c is a critical point. Then, find f′′(x)f''(x)f′′(x) and evaluate f′′(c)f''(c)f′′(c). If f′′(c)<0f''(c) < 0f′′(c)<0, then x=cx = cx=c is a local max.

How to use the second derivative to determine if x=cx = cx=c is a local min?

Find f′(x)f'(x)f′(x), set f′(c)=0f'(c) = 0f′(c)=0 to confirm c is a critical point. Then, find f′′(x)f''(x)f′′(x) and evaluate f′′(c)f''(c)f′′(c). If f′′(c)>0f''(c) > 0f′′(c)>0, then x=cx = cx=c is a local min.

How to find the global extremum given one critical point?

If the function is continuous and has only one critical point, and that point is a local extremum, then it's also the global extremum.

How to find critical points for f(x)=4sin(x)f(x) = 4sin(x)f(x)=4sin(x) on 0<x<2π0 < x < 2\pi0<x<2π?

  1. Find f′(x)=4cos(x)f'(x) = 4cos(x)f′(x)=4cos(x). 2. Set 4cos(x)=04cos(x) = 04cos(x)=0. 3. Solve for x, which gives x=π2,3π2x = \frac{\pi}{2}, \frac{3\pi}{2}x=2π​,23π​.

Given f(x)=4sin(x)f(x) = 4sin(x)f(x)=4sin(x) and critical point x=π2x = \frac{\pi}{2}x=2π​, how to determine if it's a local max or min?

  1. Find f′′(x)=−4sin(x)f''(x) = -4sin(x)f′′(x)=−4sin(x). 2. Evaluate f′′(π2)=−4f''(\frac{\pi}{2}) = -4f′′(2π​)=−4. 3. Since f′′(π2)<0f''(\frac{\pi}{2}) < 0f′′(2π​)<0, it's a local max.

How to determine the nature of critical points when f′(x)=247xf'(x) = \frac{247}{x}f′(x)=x247​?

  1. Set f′(x)=0f'(x) = 0f′(x)=0 to find critical points, but there are none. 2. Check where f′(x)f'(x)f′(x) is undefined, which is at x=0x=0x=0. 3. Since f(x)f(x)f(x) is not defined at x=0x=0x=0, there are no local extrema.

What is the formula for the second derivative?

f′′(x)=d2dx2f(x)f''(x) = \frac{d^2}{dx^2}f(x)f′′(x)=dx2d2​f(x)

How to determine critical points?

Solve f′(x)=0f'(x) = 0f′(x)=0 or find where f′(x)f'(x)f′(x) is undefined.

Second Derivative Test: Local Minimum

If f′(c)=0f'(c) = 0f′(c)=0 and f′′(c)>0f''(c) > 0f′′(c)>0, then f(c)f(c)f(c) is a local minimum.

Second Derivative Test: Local Maximum

If f′(c)=0f'(c) = 0f′(c)=0 and f′′(c)<0f''(c) < 0f′′(c)<0, then f(c)f(c)f(c) is a local maximum.

How to find the second derivative of f(x)=23x3−52x2−3xf(x) = \frac{2}{3}x^3-\frac{5}{2}x^2-3xf(x)=32​x3−25​x2−3x?

f′(x)=2x2−5x−3f'(x) = 2x^2 - 5x - 3f′(x)=2x2−5x−3, f′′(x)=4x−5f''(x) = 4x - 5f′′(x)=4x−5

What is the second derivative of f(x)=4sin(x)f(x) = 4sin(x)f(x)=4sin(x)?

f′(x)=4cos(x)f'(x) = 4cos(x)f′(x)=4cos(x), f′′(x)=−4sin(x)f''(x) = -4sin(x)f′′(x)=−4sin(x)

What is the second derivative of f(x)=247ln(x2)f(x) = 247ln(x^2)f(x)=247ln(x2)?

f′(x)=494xf'(x) = \frac{494}{x}f′(x)=x494​, f′′(x)=−494x2f''(x) = -\frac{494}{x^2}f′′(x)=−x2494​

What is the formula to check for concavity?

Check the sign of f′′(x)f''(x)f′′(x).

What is the formula to find inflection points?

Solve f′′(x)=0f''(x) = 0f′′(x)=0 or find where f′′(x)f''(x)f′′(x) is undefined.

What is the formula for the second derivative test?

  1. Find critical points using f′(x)f'(x)f′(x). 2. Plug critical points into f′′(x)f''(x)f′′(x). 3. Determine local min/max based on the sign of f′′(x)f''(x)f′′(x).

Define a critical point.

A point where f'(x) = 0 or f'(x) does not exist.

What is a local maximum?

A point where the function's value is greater than or equal to the values at all nearby points.

What is a local minimum?

A point where the function's value is less than or equal to the values at all nearby points.

Define concavity.

The direction in which a curve bends. Concave up means the curve opens upwards; concave down means it opens downwards.

What is an inflection point?

A point on a curve where the concavity changes (from up to down or vice versa).

What does the second derivative tell us?

The concavity of the function and helps determine local extrema.

Define the Second Derivative Test.

A method using the second derivative to determine whether a critical point is a local maximum or minimum.

What does f''(x) > 0 imply?

The function is concave up.

What does f''(x) < 0 imply?

The function is concave down.

What does 'inconclusive' mean in the context of the Second Derivative Test?

The test fails to determine whether the critical point is a local max, min, or neither. Further analysis is needed.