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(a) Find the intervals of increase or decrease.(b) Find the local maximum and minimum values.(c) Find the intervals of concavity and the inflection points.(d) Use the information from parts $ (a) - (c) $ to sketch the graph. Check your work with a graphing device if you have one.

$ f(\theta) = 2 \cos \theta + \cos^2 \theta $, $ 0 \leqslant \theta \leqslant 2\pi $

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Calculus 1 / AB

Calculus 2 / BC

Chapter 4

Applications of Differentiation

Section 3

How Derivatives Affect the Shape of a Graph

Derivatives

Differentiation

Volume

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(a) Find the intervals of …

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(a) Find the intervals…

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07:59

It'S a problem is so part a finding intervals of increase or decrease. So for part, a we take derivative which is equal to 2 times negative sine theta plus 2 times cosine theta times negative sine theta, which is equal to negative 2 times sine theta times 1. Plus cosine theta, if we let f prime, is equal to 0, when half theta is equal to pi or 0 or 2 pi and behalf, and the interval 0 to pi behalf of prim is negative, so philox is decreasing, as interval pi to 2 pi f, prime, Is greater than 0 point, so the function of decreasing increases part b find the local maximum and the minimum values from part a. We have at the point x, equal to pi. The function f has a local minimum value, which is equal to negative 1 part c. Find the intervals of concavity and the inflection point so first, we compute second derivative which is equal to negative 2 times cosine theta times 1 plus cosine theta plus 2 times sine theta square if we replace sine theta square by 1, minus cosine, theta square. So behalf this is equal to negative 2 times 2 cosine theta square plus cosine theta minus 1, which is equal to negative 2 times cosine theta plus 1 times 2 times, cosine theta minus 1 point. So we can find an interval from 0 to pi over 3. Second, derivative is negative to the function concaveand. The interval pi over 3 to 5 pi over 3 fagon derivative is not negative, so the function concave up and the interval 5 pi over 3 to 2 pi. The second derivative is negative to the function. Concave down d used the information from parts a to c to sketch the graph, so we can sketch the graph as follows: 0 pi to piers. This is 3 and negative 1 over half. So like this, this is a quart of the function of x and there's a local minimum value into 1 and by using some graphing device is the graph of the function is as follows: from 0 to 2 pi.

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