Find the absolute extrema of the function on the closed interval.
[tex]f(x)=\sin (x), \quad\left[\frac{3 \pi}{4}, \frac{7 \pi}{4}\right][/tex]
minimum [tex](x, y)=[/tex]
maximum [tex](x, y)=[/tex]
Answer (2)
Find the derivative of f ( x ) = sin ( x ) , which is f ′ ( x ) = cos ( x ) .
Set f ′ ( x ) = 0 to find critical points, resulting in x = 2 3 π within the interval [ 4 3 π , 4 7 π ] .
Evaluate f ( x ) at the endpoints and critical point: f ( 4 3 π ) = 2 2 , f ( 4 7 π ) = − 2 2 , and f ( 2 3 π ) = − 1 .
Identify the absolute maximum and minimum: maximum is ( 4 3 π , 2 2 ) and minimum is ( 2 3 π , − 1 ) .
Explanation
Problem Analysis We are given the function f ( x ) = s in ( x ) and the interval [ 4 3 p i , 4 7 p i ] . Our goal is to find the absolute maximum and minimum values of f ( x ) on this interval.
Finding Critical Points To find the absolute extrema, we first need to find the critical points of f ( x ) within the given interval. The derivative of f ( x ) = s in ( x ) is f ′ ( x ) = cos ( x ) .
Solving for Critical Points We set f ′ ( x ) = 0 to find the critical points: cos ( x ) = 0 . The general solutions are x = f r a c p i 2 + n p i , where n is an integer.
Identifying Critical Points in the Interval Now we need to find the critical points that lie within the interval [ 4 3 p i , 4 7 p i ] .
For n = 0 , x = f r a c p i 2 a pp ro x 1.57 , which is not in the interval. For n = 1 , x = f r a c p i 2 + p i = f r a c 3 p i 2 a pp ro x 4.71 , which is in the interval since f r a c 3 p i 4 a pp ro x 2.36 and f r a c 7 p i 4 a pp ro x 5.50 .
For n = 2 , x = f r a c p i 2 + 2 p i = f r a c 5 p i 2 a pp ro x 7.85 , which is not in the interval. Thus, the only critical point in the interval is x = f r a c 3 p i 2 .
Evaluating the Function Next, we evaluate f ( x ) at the endpoints of the interval and at the critical point: f ( 4 3 p i ) = s in ( 4 3 p i ) = f r a c s q r t 2 2 a pp ro x 0.707 f ( 4 7 p i ) = s in ( 4 7 p i ) = − f r a c s q r t 2 2 a pp ro x − 0.707 f ( 2 3 p i ) = s in ( 2 3 p i ) = − 1
Determining Extrema Comparing these values, we find that the absolute maximum is f r a c s q r t 2 2 at x = f r a c 3 p i 4 , and the absolute minimum is − 1 at x = f r a c 3 p i 2 .
Final Answer Therefore, the absolute maximum is \left( \\frac{3\\pi}{4}, \\frac{\\sqrt{2}}{2} \\right) and the absolute minimum is \left( \\frac{3\\pi}{2}, -1 \\right) .
Examples
Understanding how to find maximum and minimum values of functions is crucial in many real-world applications. For example, engineers use these techniques to optimize the design of structures, ensuring they can withstand maximum stress with minimal material. Similarly, economists use optimization to maximize profits or minimize costs. In physics, finding extrema helps determine stable states of systems, such as the lowest energy state of an atom. These mathematical tools provide a foundation for making informed decisions and achieving optimal outcomes in various fields.
The absolute maximum of the function f ( x ) = sin ( x ) on the interval [ 4 3 π , 4 7 π ] is ( 4 3 π , 2 2 ) , and the absolute minimum is ( 2 3 π , − 1 ) .
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