Locate the absolute extrema of the function f(x)=Sinpix on the closed interval [1,1/3]
y = sin (pi x)
graph it
sin pi = 0
if you go around counterclockwise
where is sin pi x= -1?
where pi x = 3 pi/2 sin ( pi x ) = -1
so min when x = 3/2
we never get to x = 1/2 so max at pi/3
when sin pi x = sin 60 deg = sqrt 3/2
To locate the absolute extrema of the function f(x) = sinpix on the closed interval [1, 1/3], we need to find the maximum and minimum values of the function within this interval.
1. First, let's calculate the values of f(x) at the endpoints of the interval:
- f(1) = sin(pi*1) = sin(pi) = 0
- f(1/3) = sin(pi*(1/3)) = sin(pi/3) = √3/2
2. Next, we need to find the critical points of the function within the interval. These are the points where the derivative of the function is zero or undefined. To find the derivative of f(x), we use the chain rule:
f'(x) = pi * cos(pi*x)
3. Setting f'(x) = 0, we have:
pi * cos(pi*x) = 0
Solving this equation, we find that cos(pi*x) = 0, which means pi * x is an odd multiple of pi/2. Hence, x = 1/2 is a critical point.
4. Now, let's check the value of f(x) at the critical point:
f(1/2) = sin(pi*(1/2)) = sin(pi/2) = 1
5. Lastly, we compare the function values at the endpoints and the critical point to determine the absolute extrema:
- f(1) = 0
- f(1/3) = √3/2
- f(1/2) = 1
Therefore, the absolute maximum value of f(x) is 1, achieved at x = 1/2, and the absolute minimum value is 0, achieved at x = 1.
To find the absolute extrema of the function f(x) = sin(πx) on the closed interval [1, 1/3], we need to locate the highest and lowest points on the interval.
1. First, let's find the critical points of the function within the given interval. Critical points occur where the derivative is either zero or undefined.
To find the derivative of f(x) = sin(πx), we can use the chain rule:
f'(x) = π * cos(πx)
Setting f'(x) equal to zero, we have:
π * cos(πx) = 0
cos(πx) = 0
Solving for x, we find that x = 1/2 is the only critical point within the given interval.
2. Now, we need to evaluate the function at the endpoints of the interval, x = 1 and x = 1/3.
f(1) = sin(π) = 0
f(1/3) = sin(π/3) = √3/2
3. Finally, we compare the values obtained at the critical point and the endpoints of the interval to determine the maximum and minimum values.
f(1) = 0 is the minimum value, and f(1/3) = √3/2 is the maximum value within the interval.
Therefore, the absolute minimum is f(1) = 0 at x = 1, and the absolute maximum is f(1/3) = √3/2 at x = 1/3.