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To determine where a function like \( g(x) = 4 \sqrt{x} - x^2 + 3 \) is increasing or decreasing, we look at its first derivative \( g'(x) \). The idea is simple:

• If \( g'(x) > 0 \), the function is increasing.
• If \( g'(x) < 0 \), the function is decreasing.

For this specific function, we found \( g'(x) = \frac{2}{\sqrt{x}} - 2x \). We check the sign of this derivative in various intervals which are determined by the critical points found from solving \( g'(x) = 0 \) and where \( g'(x) \) is undefined. In this case, critical points were \( x = 0 \) and \( x = 1 \).
Testing the intervals \((0,1)\) and \((1,\infty)\), it turns out that the function is increasing on \((0,1)\) and decreasing on \((1, \infty)\).
These intervals help us understand the behavior of the function across different sections of the domain.

Derivatives are a fundamental concept in calculus, representing the rate of change of a function. For example, the derivative of \( g(x) = 4 \sqrt{x} - x^2 + 3 \) tells us how \( g(x) \) changes for tiny changes in \( x \). Let's break it down:
Derivatives use rules like the power rule and the constant rule to simplify calculations. The rule states that for \( x^n \), the derivative is \( n \cdot x^{n-1} \).

• The derivative of \( 4 \sqrt{x} \) simplifies to \( \frac{2}{\sqrt{x}} \) using the power rule \((\sqrt{x} = x^{1/2})\).
• The derivative of \(- x^2\) is \(-2x\).
• The derivative of a constant like 3 is 0.

Bringing these components together, the derivative \( g'(x) = \frac{2}{\sqrt{x}} - 2x \) gives us a tool to analyze the function's rates of change.

Extreme values in a function refer to the highest or lowest points, either within a particular region or across the entire domain. For the function \( g(x) = 4 \sqrt{x} - x^2 + 3 \), extreme values can be categorized as local or absolute:

• A local maximum is a peak within a region: It's higher than surrounding values, but not necessarily the highest overall.
• An absolute maximum is the highest point the function reaches across its entire domain.

By evaluating the function \( g(x) \) at critical points and endpoints, we identified a local maximum of 6 at \( x = 1 \) and an absolute minimum of 3 at \( x = 0 \). Functions often reach their highest and lowest points where the derivative changes sign, just like \( g(x) \) does here.
The identification of extrema involves checking critical points and observing the behavior, ensuring no values are overlooked.
It’s a crucial step in understanding how dynamic and variable the function can be.

Critical points of a function are specific values of \( x \) where the derivative is zero or undefined. They are essential in analyzing the behavior of functions:

• Critical points are potential locations for extreme values like maxima, minima, or points of inflection.
• These points give us crucial insight into changes in the function's behavior, such as transitioning from increasing to decreasing.

For \( g(x) = 4 \sqrt{x} - x^2 + 3 \), the critical points were calculated by setting \( g'(x) = 0 \) and considering where \( g'(x) \) is undefined. In this case, they are located at \( x = 0 \) and \( x = 1 \).
In calculus, correctly identifying and evaluating critical points can reveal much about the graph of a function, highlighting key features like peaks and valleys. Critical points serve as the foundation for further analysis, such as determining increasing and decreasing intervals.