Question

F'(x) = xsin²x - 1/x is the derivative of a function f.

How many critical numbers does f have on the open interval (0, 2л)?
3
1
4
5

Answer 1

The function f'(x) = xsin²(x) - 1/x has one critical number on the open interval (0, 2π).

Step-by-step explanation:

The critical numbers of a function are the values of x where the derivative is equal to zero or is undefined. To find the critical numbers of the function f'(x) = xsin²(x) - 1/x, we need to set the derivative equal to zero and solve for x:

xsin²(x) - 1/x = 0

Multiply both sides of the equation by x:

x²sin²(x) - 1 = 0

Then use trigonometric identities to simplify the equation:

(x² - x²cos²(x)) - 1 = 0

x²(1 - cos²(x)) - 1 = 0

x²sin²(x) - 1 = 0

Therefore, there is one critical number on the open interval (0, 2π).