Absolute Extrema Of 3cos(x) On [π/2, 3π/2]
Are you grappling with finding the absolute extrema of trigonometric functions? Specifically, let's dive into the function f(x) = 3cos(x) over the interval [π/2, 3π/2]. This detailed guide will walk you through the process step by step, ensuring you understand not just the how, but also the why behind each calculation. We'll pinpoint the absolute minimum and absolute maximum values of this function within the given interval, making it easier for you to tackle similar problems in the future.
Understanding Absolute Extrema
Before we jump into the specifics of our function, let's clarify what we mean by absolute extrema. The absolute extrema of a function on a given interval are the highest and lowest points the function reaches within that interval. The absolute maximum is the largest value the function attains, while the absolute minimum is the smallest value. These points are crucial in various applications, from optimization problems in calculus to understanding the behavior of periodic functions like cosine.
To find these absolute extrema, we need to consider two types of points: critical points and endpoints of the interval. Critical points are where the derivative of the function is either zero or undefined. These points are potential locations for local maxima or minima, which might also be the absolute extrema. The endpoints, on the other hand, define the boundaries of our interval and can sometimes be where the absolute extrema occur.
Step 1: Finding Critical Points
Our first task is to find the critical points of the function f(x) = 3cos(x) on the interval [π/2, 3π/2]. To do this, we need to find the derivative of the function and set it equal to zero.
The derivative of f(x) = 3cos(x), denoted as f'(x), is found using basic trigonometric differentiation rules. The derivative of cos(x) is -sin(x), so:
f'(x) = -3sin(x)
Now, we set f'(x) equal to zero and solve for x:
-3sin(x) = 0
Dividing both sides by -3, we get:
sin(x) = 0
We need to find the values of x within our interval [π/2, 3π/2] where sin(x) equals zero. Recall that sin(x) represents the y-coordinate on the unit circle. The sine function is zero at integer multiples of π. Thus, the general solution is:
x = nπ, where n is an integer.
Within our interval [π/2, 3π/2], we have two relevant solutions:
- When n = 1, x = π
- When n = 2, x = 2π
However, 2π is outside our interval [π/2, 3π/2], so we only consider x = π as a critical point within our interval.
Step 2: Evaluating the Function at Critical Points and Endpoints
Now that we've found our critical point x = π, we need to evaluate our original function f(x) = 3cos(x) at this point, as well as at the endpoints of our interval, which are π/2 and 3π/2. This will give us the values of the function at these crucial points, allowing us to compare them and identify the absolute extrema.
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Evaluate f(x) at the critical point x = π:
f(π) = 3cos(π)
Since cos(π) = -1,
f(π) = 3(-1) = -3
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Evaluate f(x) at the endpoint x = π/2:
f(π/2) = 3cos(π/2)
Since cos(π/2) = 0,
f(π/2) = 3(0) = 0
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Evaluate f(x) at the endpoint x = 3π/2:
f(3π/2) = 3cos(3π/2)
Since cos(3π/2) = 0,
f(3π/2) = 3(0) = 0
Step 3: Identifying Absolute Extrema
We've now evaluated our function at the critical point and the endpoints. Let's summarize the values we found:
- f(π) = -3
- f(π/2) = 0
- f(3π/2) = 0
By comparing these values, we can identify the absolute maximum and absolute minimum of the function f(x) = 3cos(x) on the interval [π/2, 3π/2].
The smallest value is -3, which occurs at x = π. Therefore, the absolute minimum is -3.
The largest value is 0, which occurs at both x = π/2 and x = 3π/2. Therefore, the absolute maximum is 0.
Conclusion
In summary, for the function f(x) = 3cos(x) on the interval [π/2, 3π/2]:
- The absolute minimum is -3, occurring at x = π.
- The absolute maximum is 0, occurring at x = π/2 and x = 3π/2.
Understanding how to find absolute extrema is essential in calculus and has practical applications in optimization problems. By following these steps—finding critical points, evaluating the function at critical points and endpoints, and comparing the values—you can confidently solve similar problems.
For more information on calculus concepts and problem-solving, you might find resources on websites like Khan Academy Calculus helpful.
