Question

1)f(x) =16x^3 - 15x^2-18x+6 2)f(x)=x+2cosx Найдите точки максимума и минимума функций Помогите пожалуйста!!!

Answer 1

1)  f(x) =16x^3 - 15x^2-18x+6   
      f`(x)=48x^2-30x-18
     48x^2-30x-18=0
  D= 900-4·48·(-18)= 900+3456=4356=66^2
x1= 30+66/96=33/32
x2= 30-66/96=-3/8
 O: xmax=33/32       xmin=-3/8

2)f(x)=x+2cosx
   f`(x)= 1-2sinx
  1-2sinx=0
   -2sinx=-1
    sinx= -1/2
   x=(-1)^k · 7П/6 + Пn , n∈Z
O: xmax

Answer 2

Finding the Maximum and Minimum Points of the Functions

To find the maximum and minimum points of the given functions, we need to find the critical points where the derivative of the function is equal to zero or does not exist. Let's find the critical points for each function:

1) f(x) = 16x^3 - 15x^2 - 18x + 6:

To find the critical points, we need to find the derivative of the function and set it equal to zero:

f'(x) = 48x^2 - 30x - 18

Setting f'(x) = 0 and solving for x:

48x^2 - 30x - 18 = 0

Using the quadratic formula, we can find the values of x:

x = (-(-30) ± √((-30)^2 - 4(48)(-18))) / (2(48))

Simplifying the equation, we get:

x = (30 ± √(900 + 3456)) / 96

x = (30 ± √4356) / 96

x = (30 ± 66) / 96

So the critical points are x = 96/96 = 1 and x = -36/96 = -3/8.

Now, we need to determine whether these critical points correspond to maximum or minimum points. We can do this by analyzing the concavity of the function.

To determine the concavity, we need to find the second derivative of the function:

f''(x) = 96x - 30

Substituting the critical points into the second derivative:

f''(1) = 96(1) - 30 = 66 f''(-3/8) = 96(-3/8) - 30 = -54

Since f''(1) > 0, the point x = 1 corresponds to a minimum point, and since f''(-3/8) < 0, the point x = -3/8 corresponds to a maximum point.

Therefore, the function f(x) = 16x^3 - 15x^2 - 18x + 6 has a minimum point at x = 1 and a maximum point at x = -3/8.

2) f(x) = x + 2cos(x):

To find the critical points, we need to find the derivative of the function and set it equal to zero:

f'(x) = 1 - 2sin(x)

Setting f'(x) = 0 and solving for x:

1 - 2sin(x) = 0

2sin(x) = 1

sin(x) = 1/2

Using the inverse sine function, we can find the values of x:

x = π/6 + 2πn or x = 5π/6 + 2πn, where n is an integer.

Now, let's analyze the concavity of the function to determine whether these critical points correspond to maximum or minimum points.

To determine the concavity, we need to find the second derivative of the function:

f''(x) = -2cos(x)

Substituting the critical points into the second derivative:

f''(π/6) = -2cos(π/6) = -√3 f''(5π/6) = -2cos(5π/6) = √3

Since f''(π/6) < 0, the point x = π/6 corresponds to a maximum point, and since f''(5π/6) > 0, the point x = 5π/6 corresponds to a minimum point.

Therefore, the function f(x) = x + 2cos(x) has a maximum point at x = π/6 and a minimum point at x = 5π/6.

I hope this helps! Let me know if you have any further questions.

Answer 3

Finding Maximum and Minimum Points of the Functions

To find the maximum and minimum points of the given functions, we'll need to differentiate the functions and then solve for critical points. Let's start with the first function:

1. Function 1: - \( f(x) = 16x^3 - 15x^2 - 18x + 6 \)

To find the maximum and minimum points of this function, we'll first find its derivative and then solve for critical points.

The derivative of \( f(x) \) is: \[ f'(x) = 48x^2 - 30x - 18 \]

Setting \( f'(x) = 0 \) to find critical points: \[ 48x^2 - 30x - 18 = 0 \]

Using the quadratic formula, we can solve for \( x \) to find the critical points.

2. Function 2: - \( f(x) = x + 2\cos(x) \)

Similarly, for this function, we'll find its derivative and solve for critical points.

The derivative of \( f(x) \) is: \[ f'(x) = 1 - 2\sin(x) \]

Setting \( f'(x) = 0 \) to find critical points: \[ 1 - 2\sin(x) = 0 \]

Solving for \( x \) will give us the critical points for this function.

After finding the critical points for both functions, we can then determine which of these points correspond to maximum or minimum values by using the second derivative test or by analyzing the behavior of the functions around these points.

Let's proceed with finding the critical points for both functions and then determining the nature of these points.