Question

s(x)=(4-x)*(x√3/2) , найти наибольшее и наименьшее Ребят, помогите пожалуйста=)))

Answer 1

$y=(4-x)\cdot x\frac{\sqrt3}{2}\\\\y'=\frac{\sqrt3}{2}(-1\cdot x+(4-x)\cdot 1)=\frac{\sqrt3}{2}(-x+4-x)=\\\\=\frac{\sqrt3}{2}(4-2x)=\sqrt3(2-x)=0,\\\\2-x=0,\; x=2$

Считаем знаки производной на интервалах:  + + + + +(2) - - - - - - - 
х=2 абсцисса точки max
y(max)=y(2)=2√3 

Answer 2

Calculating the Maximum and Minimum of s(x)

To find the maximum and minimum of the function s(x) = (4-x)*(x√3/2), we can use calculus to find the critical points and determine whether they correspond to maximum or minimum values.

First, let's find the derivative of s(x) with respect to x, and then solve for the critical points by setting the derivative equal to zero.

The derivative of s(x) is given by: s'(x) = (4 - x) * (1/2 * √3) - x * (-1/2 * √3/2)

Now, let's solve for the critical points by setting the derivative equal to zero and solving for x: (4 - x) * (1/2 * √3) - x * (-1/2 * √3/2) = 0

Solving for x, we get: (4 - x) * (1/2 * √3) + x * (1/2 * √3/2) = 0 2√3 - x√3 + x√3/2 = 0 2√3 = x√3 - x√3/2 2√3 = x√3/2 4√3 = x√3 x = 4

So, the critical point is x = 4. To determine whether this critical point corresponds to a maximum or minimum, we can use the second derivative test. The second derivative of s(x) is:

s''(x) = -√3/2

Since the second derivative is negative, the critical point x = 4 corresponds to a maximum.

Therefore, the maximum value of s(x) is s(4) = (4-4)*(4√3/2) = 0.

To find the minimum value, we can check the endpoints of the domain. Since the domain of s(x) is not specified, we will assume it to be all real numbers. In this case, there are no specific endpoints, and the function does not have a minimum value.

So, the maximum value of s(x) is 0, and there is no minimum value for the function s(x).