Find the intervals on which f is increasing or decreasing and  find the local maximum and minimum values of f. Find the intervals of concavity and the inflection points. f(x)=sinx+cosx, 0<x<2pi` `

Expert Answers

An illustration of the letter 'A' in a speech bubbles

Hello! This function is infinitely differential on entire real line, therefore we can use derivatives to determine its behavior.

Note that `f(x) = sinx + cosx = sqrt(2) sin(x+pi/4).`

The first derivative is `f'(x) = sqrt(2) cos(x + pi/4).` It is zero (inside `[0, 2 pi]` ) at `x_1 =...

See
This Answer Now

Start your 48-hour free trial to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

Get 48 Hours Free Access

Hello! This function is infinitely differential on entire real line, therefore we can use derivatives to determine its behavior.

Note that `f(x) = sinx + cosx = sqrt(2) sin(x+pi/4).`

The first derivative is `f'(x) = sqrt(2) cos(x + pi/4).` It is zero (inside `[0, 2 pi]` ) at `x_1 = pi/4` and `x_2 = pi/4 + pi = (5pi)/4.`

This way the first derivative is positive (and `f(x)` increases) on `(0, pi/4)` and on `((5pi)/4, 2pi)` and is negative (and `f(x)` decreases) on `(pi/4, (5pi)/4).` Thus `f(x)` has a local maximum at `x_1 = pi/4` with the value `f(pi/4) = sqrt(2)` and the local minimum at `x_2=(5pi)/4` with the value `f(x_2)=-sqrt(2).`

The second derivative is `f''(x) = -sqrt(2) sin(x+pi/4),` and it is negative on `(0, (3pi)/4)` and on `((7pi)/4,2pi)` and is positive on `((3pi)/4,(7pi)/4).` Therefore `f(x)` is concave down on `(0, (3pi)/4)` and on `((7pi)/4,2pi)` and is concave up on `((3pi)/4,(7pi)/4).`

The inflection points are where direction of concavity changes, i.e. `(3pi)/4` and `(7pi)/4.`

Look at the picture made with desmos.com. The function is in green, the first derivative is dashed in red and the second derivative is dotted in red.

Images:
This image has been Flagged as inappropriate Click to unflag
Image (1 of 1)
Approved by eNotes Editorial Team