Question

201 А) f(x)=cos(3x^2-4x+2) Б) f(x)=sin(2x^2-3x+1) 202 А) f(x)=корень3-cos x все под корнем Б) f(x)=все под корнем5+sin x

Answer 1

1.Найдите производную функции

а)y=( 8x - 15)^5

   y`(x)=5(8x-15)^4 * 8=40(8x-15)^4

б)y=sqrt{3 - 2x}

  y`(x)=-2/(2sqrt{3-2x)}=-1/sqrt{3-2x}

в)y= sin(4x + пи/6)

   y`(x)=4cos(4x + пи/6)

г)y=1/1-3x

   y`(x)=(-1)(-3)/(1-3x)^2=3/(1-3x)^2

 

2.

Решите неравенство f'(x)<0, если f(x)=-x^3+3x^2-4

f`(x)=-3x^2+6x=-3x(x-2)

-3x(x-2)<0

 

   -              +             -

--------(0)--------(2)-------

 

(- бесконечность; 0) объединение (2; + бесконечность)

Answer 2

201 A) To find the critical points of the function f(x) = cos(3x^2 - 4x + 2), we need to find where the derivative is equal to zero:

f'(x) = -sin(3x^2 - 4x + 2) * (6x - 4)

Setting f'(x) = 0 gives:

sin(3x^2 - 4x + 2) = 0 or 6x - 4 = 0

For the first equation, we have:

3x^2 - 4x + 2 = nπ where n is an integer

Using the quadratic formula, we get:

x = (2 ± √(4 - 12(2 - nπ))) / 6

Simplifying:

x = (1 ± √(1 - 3(2 - nπ))) / 3

For the second equation, we have:

6x - 4 = 0

x = 2/3

So the critical points of the function are given by:

x = (1 ± √(1 - 3(2 - nπ))) / 3 for n an integer, and x = 2/3

To determine the nature of these critical points, we need to examine the sign of the second derivative at each point. The second derivative is:

f''(x) = -6(6x^2 - 7)sin(3x^2 - 4x + 2)

When sin(3x^2 - 4x + 2) = 0, we have:

f''(x) = -6(6x^2 - 7)sin(3x^2 - 4x + 2) = -6(6x^2 - 7) * 0 = 0

So we need to consider the points where sin(3x^2 - 4x + 2) ≠ 0. At these points, the sign of the second derivative will determine whether the critical point is a maximum or minimum.

To determine the sign of the second derivative, we need to examine the sign of 6x^2 - 7. This is positive for x < -√7/6 or x > √7/6, and negative for -√7/6 < x < √7/6. Therefore, we have:

• f''(x) > 0 for x < -√7/6 and x > √7/6 (minima)
• f''(x) < 0 for -√7/6 < x < √7/6 (maxima)

So the critical points of the function are:

• x = (1 + √(1 - 3(2 - nπ))) / 3 for n an odd integer: local minimum
• x = (1 - √(1 - 3(2 - nπ))) / 3 for n an even integer: local maximum
• x = 2/3: local minimum

201 Б) To find the critical points of the function f(x) = sin(2x^2 - 3x + 1), we need to find where the derivative is equal to zero:

f'(x) = 4x cos(2x^2 - 3x + 1) - 3cos(2x^2 - 3x + 1)

Setting f'(x) = 0 gives:

cos(2x^2 - 3x + 1) = 0 or 4x = 3cos(2