Question

F'(x) =0, если f(x) =x-lnx

Answer 1

"f(штрих)(х)" - это производная функции.

f(х)= ln(x+1)-2x

f'(x) = (ln(x+1)-2x)' = (ln(x+1))' - (2x)' =

Найдем, когда f'(x)=0:

Ответ: - 0,5

Answer 2

To find the critical points of the function f(x) = x - ln(x), we need to find the values of x where the derivative f'(x) equals zero.

First, let's find the derivative of f(x) with respect to x:

f(x) = x - ln(x)

Using the rules of differentiation, the derivative f'(x) is:

f'(x) = 1 - (1/x)

Now, we need to find the values of x where f'(x) equals zero:

1 - (1/x) = 0

To solve this equation for x, we can add 1/x to both sides:

1 = 1/x

Next, we can multiply both sides by x to get rid of the fraction:

x = 1

So, the critical point of the function f(x) = x - ln(x) occurs at x = 1.

To determine if this point is a minimum, maximum, or neither, we can analyze the second derivative f''(x).

First, let's find the second derivative of f(x):

f'(x) = 1 - (1/x)

Using the quotient rule to differentiate the second time, we get:

f''(x) = (d/dx)(1) - (d/dx)(1/x)

f''(x) = 0 + 1/x^2

Now, let's evaluate the second derivative at x = 1:

f''(1) = 1/(1^2) = 1

Since the second derivative f''(x) is positive for x = 1, it indicates that the function is concave up at that point.

Therefore, the critical point at x = 1 corresponds to a local minimum for the function f(x) = x - ln(x).