Classify the critical points of function using second derivitive test function is sinx-cosx 0<=x<=2pi?

Answer 1

First, find the critical points of the function, which are points on the graph where the first derivative vanishes or is undefined. The derivative of your function is f'(x) = cos(x) + sin(x). We are thus looking for values of x where sin(x) = -cos(x). Looking at the graphs of these functions, this happens twice in the interval [0, 2Pi]. The unit circle shows us that these are the arguments 3Pi/4 and 7Pi/4.
The second derivative of your function f''(x) is actually the negation of f(x). The first value yields a negative second derivative, so it marks the location of a local maximum. The second value yields a local minimum.
Note that the inflection points in the above posts are not critical points, the the first derivative neither vanishes (horizontal tangent line) nor becomes undefined (vertical tangent line) there.

Answer 2

Seting the first derivative to zero gives

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

This can be true only in the second quadrant, where sine is positive and cosine negative, and in the fourth quadrant, where sine is negative and cosine positive. The reference angle is pi/4, so the critical points are 3pi/4 and 7pi/4.

To classify the points, we need the second derivative:

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

At 3pi/4, this expression is negative, so the point is a maximum. At 7pi/4, the expression is positive, so the point is a minimum.

I suppose we need the inflection points also. These occur where the second derivative is zero.

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

x can be in the first quadrant, where sine and cosine are positive, and in the third quadrant, where they are both negative. The reference angle is again pi/4, so the inflection points are at pi/4 and 5pi/4.

At the maximum, the original function has the value sqrt(3). At the minimum, it has the value -sqrt(3). If you want to graph the function, you might like to know that it goes through the points (0, -1), (pi/4, 0), (pi/2, 1), (3pi/4, sqrt(3)), (pi, 1), (5pi/4, 0), (3pi/2, -1), (7pi/4, -sqrt(3)).

Answer 3

d( x(x-2)²+a million )/dx = d( x(x²-4x+4)+a million )/dx = d( x³-4x²+4x+a million )/dx = 3x²-8x+4 at the same time as 3x²-8x+4=0 the acute factors are: 3x²-8x+4=0 x=[8+-(8²-4*3*4)^¹/2] / (2*3) x=[8+-(sixty 4-40 8)^¹/2] / 6 x=[8+-16^¹/2] / 6 x=[8+-4] / 6 x1 = 2/3 ... or ... x2 = 2 => y1 = (2/3)³-4(2/3)²+4(2/3)+a million y1 = 8/27 - 4(4/9) + 8/3+ 27/27 y1 = 8/27 - 16/9 + 8/3+ 27/27 y1 = 8/27 - 40 8/27 + seventy 2/27+ 27/27 = fifty 9/27 and y2 = (2)³-4(2)²+4(2)+a million y2 = 8 - 16 + 8 + a million = a million => (2/3, fifty 9/27) and (2, a million) .:.:.:.:.:.:.:.:.:.:.:.:.:.:.:.:. 2d derivitive: d( 3x²-8x+4 )/dx = 6x-8 at the same time as x=2/3 contained in the 2d derivitive: y''= 6(2/3)-8 = 4 - 8 = -4 < 0 the point (2/3, fifty 9/27) is minimum aspect. at the same time as x=2 contained in the 2d derivitive: y''= 6(2)-8 = 12 - 8 = 4 > 0 the point (2/3, fifty 9/27) is maximum aspect. ?

Answer 4

f(x)=sinx-cosx
f`(x)=cosx+sinx
cosx+sinx=0 divide by cosx
tgx+1=0
tgx=-1 , x=3pi/4 and x=7pi/4 this is the critical points.
f"(x)=cosx-sinx
f"(3pi/4)=cos3pi/4-sin3pi/4=-sqrt(2)<0
f"(7pi/4)=cos7pi/4-sin7pi/4=sqrt(2)>0
Then x=3pi/4 is local max. and x=7pi/4 is local min.

Answer 5

f(x)=sinx-cosx
f'(x)=cosx+sinx
equating to zero
sinx=cosx
x=pi/4
f''(x)=cosx-sinx
substituting x=pi/4 f''(x)=0