Consider the following. f(t)= t + cos t, -2 pi leq t leq 2 pi (a) Find the inter
ID: 2870569 • Letter: C
Question
Consider the following. f(t)= t + cos t, -2 pi leq t leq 2 pi (a) Find the interval of increase. (Enter your answer using interval notation.) (b) Find the local maximum value(s). (Enter your answers as a comma-separated list. If an answer does not exist, enter DNE.) Find the local minimum value(s). (Enter your answers as a comma-separated list. If an answer does not exist, enter DNE.) (c) Find the inflection points. (smallest x-value) ) ) ) (largest x-value) function is concave up. (Enter your answer using interval notation.) (x, y) = ( (x, y) = ( (x, y) = ( (x, y) = ( Find the intervals the Find the intervals the function is concave down. (Enter your answer using interval notation.)Explanation / Answer
a) f(t) = t + cos(t)
Firstly we need to take the derivative of the function,
f'(t) = 1 - sin(t)
The function will take value 0, when sin(x) approaches 1 which imples x = (2n+1)pi/2 for n being even numbers
since f'(t) will always be positive hence the function will always be increasing
b) The local maximum can occur at either the points where the derivative is equal to zero or the end points
f(2pi) = 2pi + cos(2pi) = 2pi + 1
f(-2pi) = -2pi + cos(-2pi) = -2pi + 1
f(pi/2) = pi/2 + cos(pi/2) = pi/2
Hence maximum value occurs at point x=2pi and is equal to the value of 2pi + 1
c)The local minimum can occur at either the points where the derivative is equal to zero or the end points
f(2pi) = 2pi + cos(2pi) = 2pi + 1
f(-2pi) = -2pi + cos(-2pi) = -2pi + 1
f(pi/2) = pi/2 + cos(pi/2) = pi/2
Hence local minimum occurs when the function is at x=-2pi and the value is equal to 1-2pi
c) In order to find the inflection we need to find the double derivative of the function
f''(t) = -cos(t)
hence the points where f''(t) = 0 => x = pi/2,3pi/2,-pi/2,-3pi/2
First Point is -3pi/2,-3pi/2
Second Point is -pi/2,-pi/2
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