Inflection points, Intervals and minimum value of a Function

Click For Summary
SUMMARY

The discussion focuses on analyzing the function f(θ) = 2cos(θ) + (cos(θ))^2 within the interval 0 ≤ θ ≤ 2π. Participants are tasked with finding intervals of increase and decrease, local minimum values, inflection points, and concavity. Key techniques include using the first and second derivatives for sign analysis. The second derivative is essential for determining concavity and inflection points.

PREREQUISITES
  • Understanding of first and second derivatives
  • Knowledge of trigonometric functions and their properties
  • Familiarity with concavity and inflection points
  • Ability to perform sign analysis on derivatives
NEXT STEPS
  • Learn how to compute first and second derivatives of trigonometric functions
  • Study the concept of intervals of increase and decrease in calculus
  • Explore methods for finding local minima and maxima
  • Investigate concavity and inflection points in depth
USEFUL FOR

Students studying calculus, particularly those focusing on functions, derivatives, and their applications in determining function behavior.

AcecA
Messages
12
Reaction score
0

Homework Statement


Consider the function below. (Round the answers to two decimal places. If you need to use - or , enter -INFINITY or INFINITY.)
f(θ) = 2cos(θ) + (cos (θ))^2
0 ≤ θ ≤ 2π
(a) Find the interval of increase.
( , )

Find the interval of decrease.
( , )

(b) Find the local minimum value.


(c) Find the inflection points.
( , ) (smaller x value)
( , ) (larger x value)

Find the interval where the function is concave up.
( , )

Find the intervals where the function is concave down. (Enter the interval that contains smaller numbers first.)
( , ) ( , )

Homework Equations


f(θ) = 2cos(θ) + (cos (θ))2
0 ≤ θ ≤ 2π

The Attempt at a Solution


I tried using sign analysis with the second derivative (I definitely know that you have to use the second derivative) and all of the values for it. Thus:
_____________-2sinx 1 + cosx f'(x) f(x)
0 = 0
0 < x < pi/2
x = pi/2
pi/2 < x < pi
x = pi
pi < x < 3pi/2
x = 3pi/2
3pi/2 < x < 2pi
x = 2 pi

But I honestly don't know where to go from there

Homework Statement


Consider the function below. (Give your answers correct to two decimal places. If you need to use - or , enter -INFINITY or INFINITY.)
f(x) = e^[-1/(x + 2)]

(c) Find the inflection point.
( , )

Find the intervals where the function is concave up.
( , ) ( , )

Homework Equations


f(x) = e^[-1/(x + 2)]

The Attempt at a Solution


I tried something similar to the above question
 
Physics news on Phys.org
You have f(t) = 2cos(t) + cos^2(t) -- (I'm using t instead of theta)

I don't see where you have found f'(t) or f''(t). You need both of those derivatives in order to talk about where f is increasing/decreasing and where the inflection points are.
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 40 ·
2
Replies
40
Views
5K
  • · Replies 4 ·
Replies
4
Views
3K
Replies
4
Views
2K
  • · Replies 14 ·
Replies
14
Views
2K
Replies
2
Views
2K
Replies
3
Views
2K
Replies
3
Views
1K
  • · Replies 11 ·
Replies
11
Views
2K
Replies
7
Views
2K