Question

помогите с решением. не пойму что делаю не так. может с производной ошибаюсь...Найти наибольшее значение функции:1. y=9x-8sinx+7 на отрезке {-п/2;0}2. y=4tgx-4x+п-6 на отрезке { -п/4;п/4}помогите пожалуйста...

Answer 1

применены формулы производной от синуса, тангенса и линейнгой функции, а также связь поведения функции со знаком производной

Answer 2

Finding the Maximum Value of a Function

To find the maximum value of a function, we need to find the critical points of the function and determine whether they correspond to a maximum or minimum. Critical points occur where the derivative of the function is equal to zero or does not exist.

Let's solve the two problems you provided step by step:

1. Problem 1: Find the maximum value of the function y = 9x - 8sin(x) + 7 on the interval [-π/2, 0].

To find the critical points, we need to find where the derivative of the function is equal to zero or does not exist. Let's find the derivative of the function first:

   y' = 9 - 8cos(x)
   

Now, let's set the derivative equal to zero and solve for x:

   9 - 8cos(x) = 0
   

Solving this equation will give us the critical points. Let's solve it:

   8cos(x) = 9
   cos(x) = 9/8
   

Taking the inverse cosine of both sides, we get:

   x = arccos(9/8)
   

However, we need to find the critical points within the given interval [-π/2, 0]. The value of arccos(9/8) is outside this interval, so we can ignore it.

Next, we need to check the endpoints of the interval. Let's evaluate the function at x = -π/2 and x = 0:

   y(-π/2) = 9(-π/2) - 8sin(-π/2) + 7 = -9π/2 + 8 + 7 = -9π/2 + 15
   y(0) = 9(0) - 8sin(0) + 7 = 0 - 0 + 7 = 7
   

Now, we compare the values of the function at the critical points and endpoints to determine the maximum value:

- y(-π/2) = -9π/2 + 15 - y(0) = 7

Comparing these values, we can see that y(0) = 7 is the maximum value of the function on the interval [-π/2, 0].

2. Problem 2: Find the maximum value of the function y = 4tan(x) - 4x + π - 6 on the interval [-π/4, π/4].

Similarly, let's find the derivative of the function:

   y' = 4sec^2(x) - 4
   

Setting the derivative equal to zero and solving for x:

   4sec^2(x) - 4 = 0
   sec^2(x) = 1
   

Taking the square root of both sides, we get:

   sec(x) = ±1
   

However, we need to find the critical points within the given interval [-π/4, π/4]. The values of x that satisfy sec(x) = ±1 are x = -π/4 and x = π/4. Let's evaluate the function at these critical points:

   y(-π/4) = 4tan(-π/4) - 4(-π/4) + π - 6 = -4 + π + π - 6 = 2π - 10
   y(π/4) = 4tan(π/4) - 4(π/4) + π - 6 = 4 + π - π - 6 = -2
   

Now, we compare the values of the function at the critical points and endpoints to determine the maximum value:

- y(-π/4) = 2π - 10 - y(π/4) = -2

Comparing these values, we can see that y(-π/4) = 2π - 10 is the maximum value of the function on the interval [-π/4, π/4].

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