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What are all the critical points of f(x)=sinx+cosx ? 0=<x=<2pi
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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).
Posted by giorgiana1976 on February 22, 2011 at 2:49 AM (Answer #1)
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).
Posted by justaguide on February 22, 2011 at 2:55 AM (Answer #2)
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