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Critical points of a function are values of \( x \) where the derivative \( f'(x) \) is zero or undefined. These points are significant because they indicate potential locations of relative extrema, which can be either maximum or minimum values on the function's graph.

In our example, the function is \( f(x) = 2x + 3x^{2/3} \). We calculate its derivative to find the critical points. By setting the derivative \( f'(x) = 2 + 2x^{-1/3} \) to zero, we obtain:

• \( 2 + 2x^{-1/3} = 0 \)
• Solving yields \( x = -1 \)

Notice that if the derivative is undefined at some \( x \), that point is a critical point too, but in this scenario, \( x = -1 \) is the primary focus. Understanding critical points helps us explore relative maxima and minima in functions.

The derivative of a function provides us with crucial information about its behaviour regarding slope, direction, and rate of change. It tells us how the function's value changes as \( x \) changes. To compute the derivative of a function like \( f(x) = 2x + 3x^{2/3} \):

• Apply basic differentiation rules (power rule and sum rule)
• For \( f(x) = 2x + 3x^{2/3} \), the derivative is \( f'(x) = 2 + 2x^{-1/3} \)

Finding the derivative is the first step when searching for critical points and analyzing a function's growth or decay. It's about assessing the instantaneous rate of change at each point along the graph of the function.

The second derivative test is a valuable technique to determine the nature (maximum or minimum) of critical points in a function. After finding the critical points by setting the first derivative to zero or identifying where it's undefined, the second derivative helps determine if these points are maxima, minima, or points of inflection.

To apply the second derivative test: 1. Calculate the second derivative of the function. For our example, \( f''(x) = -\frac{2}{3} x^{-4/3} \).2. Evaluate the second derivative at the critical point. Here, at \( x = -1 \), \( f''(-1) = -\frac{2}{3} \).3. Interpret the result:

• If \( f''(x) > 0 \), the function is concave up, and there's a relative minimum.
• If \( f''(x) < 0 \), the function is concave down, and there's a relative maximum.
• If \( f''(x) = 0 \), the test is inconclusive.

In this scenario, since \( f''(-1) < 0 \), the function has a relative maximum at \( x = -1 \). The second derivative test provides insights into the curvature of the function at critical points, further refining our understanding of its behavior.