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

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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 The function is in green, the first derivative is dashed in red and the second derivative is dotted in red.

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