# Inflection points, Intervals and minimum value of a Function

## 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

## Answers and Replies

Mark44
Mentor
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.