What are the answers of this question

Question

Answer the following questions about the function whose derivative is given below.

​(a) What are the critical points of​ f?

​(b) On what open intervals is f increasing or​ decreasing?

​(c) At what​ points, if​ any, does f assume local maximum and minimum​ values?

f'(x)=(√3 sinx - cosx)(sinx+√3cosx) 0<=x<=2pi

Answer 1

To find the critical points of f, one must take the derivative and set it = 0, then solve. The solution x-values are your critical numbers.

√3 sinx - cosx = 0

√3 = cosx/sinx

So tan x = 1/√3

And this is true at x = π/6 and x = 7π/6

sinx + √3 cosx = 0

√3 = -sinx/cosx

So tan x = -√3

And this is true at x = 4π/3 and x = 5π/3

0----π/6---------7π/6-------4π/3-----5π/3-----2π

Check each of these intervals to see if the derivative is positive or negative.

If the derivative is positive, then the function is increasing in that interval.

If the derivative is negative, then the function is decreasing in that interval.

(0,π/6) f ' is neg so function is decreasing

(π/6, 7π/6) f ' is pos so function is increasing

(7π/6, 4π/3) f ' is neg so function is decreasing

(4π/3, 5π/3) f ' is pos so function is increasing

(5π/3, 2π) f ' is neg so function is decreasing.

0- - - - π/6 + + + + + 7π/6 - - - - 4π/3 + + + 5π/3 - - - 2π

When the function changes from being decreasing to increasing, like at π/6 and at 4π/3 (picture it coming down then going up) there exists a minimum at each of those x-coordinates.

When the function changes from being increasing to decreasing, like at 7π/6 and at 5π/3 (picture it going up then going back down) there exists a maximum at each of those x-coordinates.

Answer 2

sqrt(3)*sin x - cosx = 0 or sinx + sqrt(3)*cosx = 0

1st equation:

sqrt(3) = cotx

sqrt(3)/3 = tan x

x = pi/6 or x = 4*pi/3

2nd equation: -sqrt(3) = tan x ---> 120 degrees = 2*pi/3 and 300 degrees = 5*pi/3

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graph the function of this derivative, so you can easily see the answers to these questions