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Locate the absolute extrema of the function on the closed interval. $$ f(x)=\cos \pi x,\left[0, \frac{1}{6}\right] $$

Short Answer

Expert verified
The absolute maximum of \(f(x) = \cos (\pi x)\) on the interval [0, 1鈦6] is \(1\) at \(x = 0\), and the absolute minimum is \(\frac{\sqrt{3}}{2}\) at \(x = \frac{1}{6}\)

Step by step solution

01

Find Critical Points

A critical point of a function occurs where the first derivative is zero or undefined. The derivative of \(f(x) = \cos (\pi x)\) is \(f'(x) = -\pi \sin (\pi x)\). The derivative does not exist for \(x=0\) and \(x=\frac{1}{6}\), and \(f'(x) = 0\) yields no solutions within the interval [0, 1鈦6], hence there are no internal critical points and the extrema only lie at the boundaries.
02

Evaluate Boundaries

Now, calculate \(f(x)\) at the boundaries of the interval. For \(x=0\), we have \(f(0) = \cos (0) = 1\). For \(x=\frac{1}{6}\), we have \(f(\frac{1}{6}) = \cos (\frac{\pi}{6}) = \frac{\sqrt{3}}{2}\)
03

Compare Points

Compare \(f(0)\) and \(f(\frac{1}{6})\). Since \(1 > \frac{\sqrt{3}}{2}\), the absolute maximum is at \(x = 0\) and the absolute minimum is at \(x = \frac{1}{6}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Critical Points of a Function
When you're searching for the high and low points of a function's graph, you're looking for what's called the critical points. A critical point occurs where the function's first derivative is either zero or undefined. This indicates a potential maximum, minimum, or point of inflection.

In mathematical terms, if you have a function denoted as f(x), you find its critical points by taking the derivative, f'(x), and setting it to zero to solve for x. If f'(x) doesn't exist at certain x-values within the domain, those are critical points as well. But remember, not all critical points correspond to an extremum, which is why further analysis, like the first derivative test, is usually required.

For instance, with the function given as f(x) = cos(蟺x), you would calculate its derivative as f'(x) = -蟺 sin(蟺x). In our case, since the derivative never equals zero on the specified interval and is never undefined, we find no critical points inside the interval. This lack of internal critical points hints that the extrema might be located at the boundary points of the interval.
First Derivative Test
The First Derivative Test is essentially a 'detective tool' for finding local maxima and minima of a function based on its first derivative. After finding the critical points of a function, this test allows you to determine whether each point is a local maximum, local minimum, or neither.

How does it work? You derive the function and analyze the sign of the derivative before and after each critical point. If the derivative changes from positive to negative at a point, it's a local maximum. Conversely, if it changes from negative to positive at the critical point, we have a local minimum. If there's no change in the sign, then the point is not an extremum.

However, for the function f(x) = cos(蟺x) on the interval [0, 1/6], since there were no critical points within the interval, the First Derivative Test doesn't provide additional information regarding the extrema within the interval. Yet, it remains an important concept for various functions with critical points within the domain of interest.
Closed Interval Analysis
Sometimes the extrema of a function occur not in the middle of the graph but at the 'ends'鈥攁nd that鈥檚 what Closed Interval Analysis is all about. This method involves analyzing functions over closed intervals [a, b], which include their endpoint values.

Here's how you do it:
  • Find all critical points within the interval.
  • Evaluate the function at each critical point.
  • Evaluate the function at the endpoint values of the interval.
  • Compare all these values to determine the absolute (global) maximum and minimum.
For the cos(蟺x) function on the interval [0, 1/6], we check the function's value at endpoints because no internal critical points were found. At the left endpoint, x=0, the function has a value of 1, and at the right endpoint, x=1/6, the value is 鈭3/2. So, in this case, applying closed interval analysis concludes that the function reaches its absolute maximum at x=0 and absolute minimum at x=1/6.

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Most popular questions from this chapter

In Exercises 71 and \(72,\) let \(f\) and \(g\) represent differentiable functions such that \(f^{\prime \prime} \neq 0\) and \(g^{\prime \prime} \neq 0\). Show that if \(f\) and \(g\) are concave upward on the interval \((a, b)\), then \(f+g\) is also concave upward on \((a, b)\).

Determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false. The graph of \(f(x)=1 / x\) is concave downward for \(x<0\) and concave upward for \(x>0\), and thus it has a point of inflection at \(x=0\)

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