# What are all the critical points of f(x)=sinx+cosx ? 0=<x=<2pi

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### 2 Answers

The critical points are determined by differentiating the function and equating the derivative to 0. It is solved to determine x.

f(x) = sin x + cos x

f'(x) = cos x - sin x = 0

=> cos x = sin x

=> tan x = 1

=> x = arc tan 1

=> x = pi/4 , 5*pi/4

At x = pi/4 , f(x) = sqrt 2

at x = 5*pi/4, f(x) = -sqrt 2

**The critical points are at x = pi/4 and x = 5*pi/4, and the extreme values are (pi/4, sqrt 2) and (5*pi/4,-sqrt 2).**

The critical points of a function are the roots of the first derivative of that function.

We'll have to determine the first derivative of the given function:

f'(x) = (sin x + cos x)'

f'(x) = cos x - sin x

We'll put f'(x) = 0.

cos x - sin x = 0

cos x = sin x

We'll divide by cos x both sides and we'll get:

sin x/cos x = 1

tan x = 1

The tangent has the positive value 1 in the 1st and 3rd quadrants.

x = pi/4 (1st quadrant)

x = pi + pi/4

x = 5pi/4 (3rd quadrant)

The extreme points of the function are:

f(pi/4) = sin pi/4 + cos pi/4 = 2sqrt2/2 = sqrt2

f(5pi/4) = sin pi/4 + cos pi/4 = -2sqrt2/2 = -sqrt2

**The critical points of the function are x = pi/4 and x = 5pi/4 and the extreme points are (pi/4 ; sqrt2) and (5pi/4 ; -sqrt2).**