Given the following graph of
𝑓
(
𝜃
)
=
cos
⁡
(
𝜃
)
, at what angle measure do the output values of
𝑓
change from increasing at an increasing rate to increasing at a decreasing rate?

Question

Given the following graph of 
𝑓
(
𝜃
)
=
cos
⁡
(
𝜃
)
, at what angle measure do the output values of 
𝑓
 change from increasing at an increasing rate to increasing at a decreasing rate?

BlackTom AI · Accepted Answer

We are given f(theta) = cos(theta) and asked to identify the angle where the output values change from increasing at an increasing rate to increasing at a decreasing rate.

Option 1: pi. At theta = pi, cos(pi) = -1 and the derivative f'(theta) = -sin(theta) equals 0. Since f' changes sign around pi, this is not the inflection point where the rate of increase switches from accelerating to decelerating while the function is still increasing. In fact, near theta = pi the function is decreasing, not increasing, so this cannot mark the transition from increasing-at-increasing-rate to increasing-at-decreasing-rate.

Option 2: 2*pi. At theta = 2*pi, cos(2*pi) = 1 and f'(theta) = -sin(2*pi) = 0. Just like at theta = 0, the derivative is zero and the function is transitioning locally, but the slope around 2*pi indicates a return to the same behavior as at 0, where the function is not in the regime of increasing-at-increasing-rate transitioning to increasing-at-decreasing-rate. This point is not the inflection where the concavity changes while the function is increasing.

Option 3: pi/2. At theta = pi/2, cos(pi/2) = 0 and f'(theta) = -sin(pi/2) = -1, which is negative. The function is decreasing there, not increasing. Since the question concerns the transition while the function is increasing, pi/2 is not the relevant inflection point for the described transition.

Option 4: 0. At theta = 0, cos(0) = 1 and f'(theta) = -sin(0) = 0. Right around 0, for small positive angles, sin(theta) is positive, so f' is negative and the function is decreasing, not increasing. Thus this point does not mark the switch from increasing-at-increasing-rate to increasing-at-decreasing-rate.

Option 5: 3*pi/2. At theta = 3*pi/2, cos(3*pi/2) = 0 and f'(theta) = -sin(3*pi/2) = 1, which means the function is increasing there. The second derivative f''(theta) = -cos(theta) is zero at this theta (since cos(3*pi/2) = 0), indicating an inflection point where concavity changes. To the left of 3*pi/2, f''(theta) > 0 or < 0 depending on the exact region, but right at the inflection point the concavity changes, corresponding to the transition from increasing at an increasing rate to increasing at a decreasing rate when the function is increasing.

In summary, the inflection point where the function is increasing and the concavity changes (so the rate of increase switches from accelerating to decelerating) occurs at theta = 3*pi/2. The other options either correspond to points where the function is decreasing, have zero slope without the required concavity transition while increasing, or do not represent a change in the increasing-rate behavior.

MATH 1151 AU2025 (18443) Calculus Concept Readiness Assessment (CCR)

Given the following graph of 𝑓 ( 𝜃 ) = cos ⁡ ( 𝜃 ) , at what angle measure do the output values of 𝑓 change from increasing at an increasing rate to increasing at a decreasing rate?

查看解析

登录即可查看完整答案

类似问题

For the function f(x) f(x) , if the point of inflection is located at (a,b) (a,b) , find the value of a a .f(x)=x3+−5x2+−9x+45 f(x)=x^3+-5x^2+-9x+45 Calculate your answer correct to 2 decimal places.

For the function [math: f(x)] , if the point of inflection is located at [math: (a,b)] , find the value of [math: a] .[math: f(x)=x3+−5x2+−9x+45] f(x)=x^3+-5x^2+-9x+45 Calculate your answer correct to 2 decimal places.

Question textThe function f \colon \mathbb{R} \rightarrow \mathbb{R}, \quad f(x) = x^2 e^{-x} has two inflection points. These occur when x = a\pm\sqrt{b} for non-negative integer values a= Answer 1 Question 26[input] and b= Answer 2 Question 26[input].

Question textThe function f \colon \mathbb{R} \rightarrow \mathbb{R}, \quad f(x) = x^2 e^{-x} has two inflection points. These occur when x = a\pm\sqrt{b} for non-negative integer values a= Answer 1 Question 26[input] and b= Answer 2 Question 26[input].

This is the graph of the derivative of 𝑓 . What are the x-values of the inflection points for 𝑓 ?

Suppose 𝑔 ′ ( 𝑥 ) = 𝑥 𝑥 2 + 1 . Where are the inflection points for 𝑔 ( 𝑥 ) ?

Suppose . Where are the inflection points for ?

Let be a continuous function and let be in the domain of . Suppose that when and that when . Then is a point of inflection .

更多留学生实用工具

考试浏览器助手

风格化写作助手

论文查重助手

文献引用助手

课堂转译助手

课堂笔记助手

Quiz搜索助手

学校历年真题

智能刷题助手

智能匹配练习

希望你的学习变得更简单