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

Expert Answers info

justaguide eNotes educator | Certified Educator

calendarEducator since 2010

write12,544 answers

starTop subjects are Math, Science, and Business

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).

check Approved by eNotes Editorial


giorgiana1976 | Student

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).

check Approved by eNotes Editorial