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One of the foundational concepts in calculus is the derivative. The derivative of a function gives us the slope of the function's curve at any given point. For the function \( f(x)=4x+\frac{1}{x} \), finding the derivative involves applying basic differentiation rules.

Let's do a recap of the rules:

• The power rule states that the derivative of \( x^n \) is \( nx^{n-1} \).
• The sum rule indicates that the derivative of the sum of two functions is the sum of their derivatives.
• The derivative of a constant times a function is the constant times the derivative of the function.
• The derivative of \( \frac{1}{x} \) (or \( x^{-1} \)) is \( -x^{-2} \).

Applying these rules gives \( f'(x) = 4 - \frac{1}{x^2} \). This result represents the slopes of the tangent lines at any point on the curve of \( f(x) \). If you understand these derivative rules, tackling more complex functions becomes much easier.

In calculus, identifying critical points is crucial since they can help determine where a function's maxima and minima may occur. Critical points occur where the derivative is zero or undefined.

For our function, \( f'(x) = 4 - \frac{1}{x^2} \), you find critical points by setting the derivative equal to zero. Solving \( 4 - \frac{1}{x^2} = 0 \) gives you \( x^2 = \frac{1}{4} \), which yields two solutions: \( x = \pm\frac{1}{2} \).

However, only points within the interval \([1, 3]\) are considered relevant here.

This limits our critical points to those within the specified range of the function.

To determine the absolute maximum and minimum, we evaluate the function at critical points and at the endpoints of the given interval. This approach ensures we cover all possible heights of the function's graph within the interval.

Evaluating \( f(x) \) at the endpoints \( x=1 \) and \( x=3 \), as well as the critical point within the interval, helps determine the absolute extrema.

Plugging these values in:

• \( f(1) = 4 \times 1 + \frac{1}{1} = 5 \)
• \( f\left(\frac{1}{2}\right) = 4\left(\frac{1}{2}\right) + \frac{1}{\frac{1}{2}} = 4 \)
• \( f(3) = 4 \times 3 + \frac{1}{3} = 12 + \frac{1}{3} \)

Comparing the results shows the function's peak at \( x=3 \) is \( 12\frac{1}{3} \), its maximum, and \( x=\frac{1}{2} \) is its minimum, equaling 4.

Interval evaluation is a key tactic for understanding the behavior of functions over specific domains. By examining the function within a given range, like \([1, 3]\), we can accurately determine its gains and losses, changes in incline, and peaks or troughs.

The interval shapes where critical points apply and which endpoints to assess. This is critical when defining absolute minimums or maximums since only certain portions of the function's natural domain are in play.

In this case, after calculating and observing function values at the interval's endpoints and any other notable points or potential extrema within that range, you can securely ascertain the function's overall behavior within the established interval.