If f(x)=sinxcosx, what are the critical points? on the interval (o, 2pi)
To find the critical points of the function f(x) = sin(x) cos(x) on the interval (0, 2π), we need to identify the values of x at which the derivative of f(x) is equal to zero or does not exist.
First, let's find the derivative of f(x) with respect to x using the product rule:
f'(x) = (cos^2(x) - sin^2(x))
To find the critical points, we set the derivative f'(x) equal to zero:
cos^2(x) - sin^2(x) = 0
Next, we can use the Pythagorean identity sin^2(x) + cos^2(x) = 1 to simplify the equation:
cos^2(x) - (1 - cos^2(x)) = 0
Simplifying further, we have:
2cos^2(x) - 1 = 0
Now, we solve for cos^2(x):
cos^2(x) = 1/2
Taking the square root of both sides, we get:
cos(x) = ± √(1/2)
Now, we need to find the values of x between 0 and 2π that satisfy this equation.
For cos(x) = √(1/2), x can be π/4 or 7π/4.
For cos(x) = -√(1/2), x can be 3π/4 or 5π/4.
So, the critical points of f(x) = sin(x) cos(x) on the interval (0, 2π) are: π/4, 3π/4, 5π/4, and 7π/4.
Critical points are points in the given interval (0,2π) at which the derivative is zero, or the function is not differentiable.
The function f(x)=sin(x)cos(x) is differentiable from (-∞,∞), so you only have to look at the points at which the derivative is zero.