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In calculus, the first derivative of a function provides us with valuable insights into the behavior of the function itself. When we take the first derivative of a function like \( g(x) = 4\sqrt{x} - x^2 + 3 \), it helps us understand how the function is changing at any given point. The derivative, denoted as \( g'(x) \), is essentially the slope of the tangent line to the function at any point \( x \). To calculate the first derivative, apply the power rule and chain rule. For \( g(x) \), we differentiate each term individually:

• For \( 4\sqrt{x} \), rewrite it as \( 4x^{1/2} \) and differentiate to get \( \frac{2}{\sqrt{x}} \).
• For \( -x^2 \), apply the power rule to obtain \(-2x \).
• The derivative of a constant, like 3, is zero.

Putting it all together, the derivative is \( g'(x) = \frac{2}{\sqrt{x}} - 2x \). This tells us where the function is increasing or decreasing depending on the sign of \( g'(x) \). When the derivative is positive, the function is increasing; when negative, it's decreasing. Understanding this is key to analyzing the behavior of the function over its domain.

Critical points of a function occur where its first derivative \( g'(x) \) is either zero or undefined. These points are crucial as they are potential locations for local maxima or minima, points where the function changes from increasing to decreasing, or vice versa. Finding critical points involves solving \( g'(x) = 0 \). For the function \( g(x) = 4\sqrt{x} - x^2 + 3 \), the critical points are determined by solving \( \frac{2}{\sqrt{x}} - 2x = 0 \). This can be simplified to \( \frac{2}{\sqrt{x}} = 2x \), leading to \( x = 1 \). Additionally, the derivative is undefined at \( x = 0 \) due to the square root in the denominator. The critical points are hence \( x = 0 \) and \( x = 1 \). These points should be evaluated further by testing intervals around them to understand how the function behaves in different sections of the domain.

Extreme values of a function are points where the function reaches a local maximum or minimum. These are explicitly tied to its critical points. To determine these, evaluate the function \( g(x) \) at each critical point and use limits to understand behavior at infinity. For our function, evaluate \( g(x) \) at the critical points:

• At \( x = 0 \), \( g(0) = 3 \).
• At \( x = 1 \), \( g(1) = 6 \).

The function's behavior as \( x \to \infty \) suggests \( g(x) \to -\infty \), indicating no absolute maximum in the entire domain. However, \( g(x) \) does have a local maximum at \( x = 1 \) where the function's value is 6 since the function switches from increasing to decreasing at this point.Identifying these extreme values not only helps in understanding the peaks and valleys of the graph but also allows us to examine the overall shape and tendencies of a function. This analysis is a fundamental technique in calculus, particularly in applications involving optimization.