Question

Find, if possible, the (global) maximum and minimum values of the given function on the indicated interval. $$ F(x)=6 \sqrt{x}-4 x \text { on }[0,4] $$

Short answer

The global maximum is 0, and the global minimum is -4.

Step-by-step solution

Understand the function

The given function is \( F(x) = 6\sqrt{x} - 4x \) and we need to find its global maximum and minimum on the closed interval \([0, 4]\). This is a continuous function over the given domain and involves both a square root and linear component.

Find the derivative

To determine the critical points, we first need the derivative. The derivative of \( F(x) \) is \( F'(x) = \frac{d}{dx}(6\sqrt{x}) - \frac{d}{dx}(4x) = \frac{3}{\sqrt{x}} - 4 \).

Find critical points

Set the derivative equal to zero to find critical points: \( \frac{3}{\sqrt{x}} - 4 = 0 \). Solving gives \( \frac{3}{\sqrt{x}} = 4 \) which implies \( \sqrt{x} = \frac{3}{4} \). Squaring both sides, \( x = \left(\frac{3}{4}\right)^2 = \frac{9}{16} \). This is a critical point since it lies in the interval \([0, 4]\).

Evaluate the function at critical points and endpoints

Evaluate \( F(x) \) at the critical point and endpoints of the interval. At \( x = 0 \), \( F(0) = 6\sqrt{0} - 4\times0 = 0 \). At \( x = 4 \), \( F(4) = 6\sqrt{4} - 4\times4 = 12 - 16 = -4 \). At \( x = \frac{9}{16} \), \( F\left(\frac{9}{16}\right) = 6\sqrt{\frac{9}{16}} - 4\times\frac{9}{16} = 6\times\frac{3}{4} - \frac{36}{16} = \frac{18}{4} - \frac{36}{16} = \frac{36}{16} - \frac{36}{16} = 0 \).

Identify the global maximum and minimum

Compare the values obtained: \( F(0) = 0 \), \( F(4) = -4 \), and \( F\left(\frac{9}{16}\right) = 0 \). The maximum value of \( F(x) \) on \([0, 4]\) is 0 (achieved at \( x = 0 \) and \( x = \frac{9}{16} \)), and the minimum value is -4 (achieved at \( x = 4 \)).

Key concepts

Critical Points

Critical points are essential in calculus when identifying where a function increases, decreases, or has potential extremities, like maxima or minima. To find these points, we first find the derivative of the given function and then equate it to zero. This is because critical points occur where the derivative is zero or undefined.
In simpler terms, critical points are where the function's slope levels out, forming a peak, valley, or a point of inflection. Such points help in analyzing the behavior of the function and are integral in solving optimization problems.

• Critical points can be calculated by setting the derivative equal to zero.
• They help in identifying turning points or changes in the graph's direction.

Calculating these can guide us uncovering the global maxima and minima within a specific interval.

Derivative

A derivative is a fundamental concept in calculus which measures how a function changes as its input varies. In simple terms, it's the function’s slope at a particular point. As we explore global maxima and minima, derivatives play a crucial role.
To find the derivative of a function, like in our case of 6√x - 4x, we differentiate it piece by piece. For 6√x, applying the power rule yields \(\frac{3}{\sqrt{x}}\); for -4x, this is simply -4. Hence, the overall derivative is \(F'(x) = \frac{3}{\sqrt{x}} - 4\).
This process leads us to find where the slope becomes zero or ceases to exist, highlighting critical points which help determine the highest or lowest values of the function:

• Derivatives help determine the rate of change.
• They are used to identify critical points by setting them to zero.

Understanding derivatives and their application is key to mastering calculus.

Continuous Function

A continuous function is one that has no interruptions in its graph; this means it flows smoothly without any breaks, holes, or jumps.
For the function \(F(x)=6\sqrt{x}-4x\) on the interval \([0,4]\), it is continuous because both the square root and linear components are continuous on this domain. This ensures that our calculations make sense throughout the interval and do not produce any undefined points or sudden changes.
Continuous functions allow us to apply calculus operations, like finding derivatives and critical points, confidently:

• Continuity ensures a function behaves predictably.
• It guarantees that we can find maximum and minimum values on closed intervals.

Understanding continuity is essential when dealing with real-world scenarios where smooth changes are expected.

Closed Interval

A closed interval in mathematics is an interval that includes its endpoints, denoted as [a, b]. In solving for global maximum and minimum values, focusing on closed intervals ensures all possible points in the range are considered.
In the example \([0,4]\), both 0 and 4 are included, making them endpoints essential to evaluate alongside any critical points identified in the interval. Evaluating the function at both critical points and endpoints guarantees we understand the entire range of the function’s behavior.
For optimization problems, a closed interval:

• Ensures the range is complete and contains endpoints.
• Supports finding absolute extrema through tested boundaries and midpoints.

Recognizing closed intervals helps make precise evaluations within set boundaries for maximal and minimal values.