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(a) Find the intervals of increase or decrease. …

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Problem 47 Hard Difficulty

(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

Calculus: Early Transcendentals

Chapter 4

Applications of Differentiation

Section 3

How Derivatives Affect the Shape of a Graph

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Video Transcript

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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