Question

Explain why or why not Determine whether the following statements are true and give an explanation or counterexample. a. The function \(f(x)=\sqrt{x}\) has a local maximum on the interval \([0, \infty)\) b. If a function has an absolute maximum on a closed interval, then the function must be continuous on that interval. c. A function \(f\) has the property that \(f^{\prime}(2)=0 .\) Therefore, \(f\) has a local extreme value at \(x=2\) d. Absolute extreme values of a function on a closed interval always occur at a critical point or an endpoint of the interval.

Short answer

Answer: Yes, the absolute extreme values of a function on a closed interval always occur at a critical point or an endpoint of the interval.

Step-by-step solution

a. Function \(f(x)=\sqrt{x}\) having a local maximum on the interval `[0, ∞)`

We must determine if \(f(x) = \sqrt{x}\) has a local maximum on the interval \([0,\infty)\). The derivative of \(f(x)\) is \(f'(x) = \frac{1}{2\sqrt{x}}\). Since the derivative is always positive for \(x > 0\), the function is always increasing on the interval \((0, \infty)\). As a result, the function \(f(x) = \sqrt{x}\) does not have a local maximum on the interval \([0, \infty)\).

b. Absolute maximum on a closed interval implying function continuity

If a function has an absolute maximum on a closed interval, then it must also have a maximum value. According to the Extreme Value Theorem, a continuous function on a closed interval must attain its maximum and minimum values. Therefore, if a function has an absolute maximum on a closed interval, it is required to be continuous on that interval. The statement is true.

c. Function \(f\) with \(f'(2) = 0\) having a local extreme value at \(x=2\)

The statement says that if \(f'(2) = 0\), then \(f\) must have a local extreme value at \(x=2\). This statement is not always true. A function can have a critical point where \(f'(x)=0\) without having a local extreme value. As a counterexample, consider \(f(x) = x^3\). The derivative \(f'(x) = 3x^2\), and \(f'(0) = 0\). However, \(f(x) = x^3\) does not have a local extreme value at \(x = 0\) because the function is neither locally increasing nor decreasing around that point.

d. Absolute extreme values occurring at a critical point or an endpoint

We need to determine if the absolute extreme values of a function on a closed interval always occur at a critical point or an endpoint of the interval. According to the Closed Interval Method, we must evaluate the function at its critical points and endpoints and compare the values to find the maximum and minimum values. Hence, the absolute extreme values of a function on a closed interval always occur at a critical point or an endpoint of the interval. This statement is true.

Key concepts

Local Maximum

A local maximum is a point on a function where the function's value is higher than the values at other nearby points. Imagine a hill within a landscape; if you're standing on top of that hill, then you are at a local maximum. In mathematical terms, for a point \(x = c\) to be a local maximum, there must exist an interval around \(c\) such that for every point \(x\) within this interval, the value \(f(c)\) is greater than or equal to \(f(x)\).

When evaluating whether a function like \(f(x) = \sqrt{x}\) has a local maximum on an interval, it's crucial to check the derivative \(f'(x)\). If \(f'(x) > 0\) over the interval, the function is increasing there, suggesting no local maximum can exist.

• The derivative \(f'(x) = \frac{1}{2\sqrt{x}}\) is always positive for \(x > 0\), indicating \(f(x) = \sqrt{x}\) is always increasing.
• Thus, \(f(x) = \sqrt{x}\) has no local maximum on \([0, \infty)\).

Absolute Maximum

An absolute maximum is the highest point over the entire domain of a function. It's like the peak of the highest mountaintop—no other point exceeds it in height. For a function to have an absolute maximum on a closed interval means that there exists some \(a\) within the interval where \(f(a)\) is greater than or equal to all other function values including at boundaries.

The Extreme Value Theorem plays a vital role here. This theorem ensures that if a function is continuous over a closed interval, it must have both an absolute maximum and minimum within that interval. Discontinuity can disrupt this guarantee.

• For example, a continuous function \(f\) on a closed interval \([a, b]\) must have at least one point where it reaches an absolute maximum, adhering to the theorem.

Critical Point

Critical points are where the function's derivative equals zero or is undefined, like flat spots on a curve. These points are significant because they *could* be where local maxima or minima occur, akin to the peaks and valleys of a roller coaster.

However, just having \(f'(x) = 0\) at point \(x = c\) does not guarantee a local maximum or minimum exists at this point. The shape of the curve around the critical point matters.

• Consider \(f(x) = x^3\) with \(f'(x) = 3x^2\): it has \(f'(0) = 0\), but is neither a maximum nor a minimum at \(x = 0\).

Thus, while critical points may signal potential extremities, further analysis is necessary to confirm their nature.

Extreme Value Theorem

The Extreme Value Theorem is a critical concept ensuring that certain functions possess both maximum and minimum values over closed intervals. Imagine you're searching for the tallest tree in a forest bordered by a fence; this theorem assures you will find it, as long as the ground (representing the function) is continuous without breaks.

This theorem applies specifically to functions that are continuous on closed intervals like \([a, b]\). If these conditions are met, the theorem guarantees both an absolute maximum and minimum exist somewhere in the interval.

• This implies that the search for extreme values is fruitful provided the function remains unbroken across its domain.
• It helps in finding extreme values by validating the continuity and closed interval criteria first.

In practice, this means evaluating a function at endpoints and any critical points within to pinpoint these extremes effectively.