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Local extrema are the highest or lowest points in a small neighborhood of a function's graph. When trying to find local extrema, the first step is to calculate the first derivative of the function. For our function, we have \[ f(x) = 8x^{1/3} + x^{4/3} \]. After finding the first derivative, \( f'(x) = \frac{8}{3}x^{-2/3} + \frac{4}{3}x^{1/3} \), we set it equal to zero to identify critical points that could be candidates for local extrema. Our critical point calculation gives us \( x = -2 \). To decide whether this point is a local maximum or minimum, we move to the second derivative test.This test tells us the concavity of the function at \( x = -2 \) by examining the sign of the second derivative there. If \( f''(x) > 0 \), the function is concave upward and \( x = -2 \) is a local minimum. In contrast, if \( f''(x) < 0 \), the function is concave downward, indicating a local maximum. For our function, \( f''(-2) > 0 \), confirming \( x = -2 \) as a local minimum.

Concavity is a measure of the direction in which a curve is bending. It can show if the graph of a function curls upwards or downwards. To analyze concavity, we use the second derivative of the function, \( f''(x) \). For our function: \[ f''(x) = -\frac{16}{9}x^{-5/3} + \frac{4}{9}x^{-2/3} \]. If \( f''(x) > 0 \), the graph is concave upward, making it look like a cup. Conversely, if \( f''(x) < 0 \), the graph is concave downward, resembling an upside-down cup.By examining the signs of \( f''(x) \) in different intervals, we see that:

• For \( x < 0 \), the second derivative is positive, indicating the function is concave upward on this interval.
• For \( x > 0 \), the second derivative becomes negative, showing a concave downward behavior.

This shift in concavity around \( x=0 \) gives us important insights into the shape of the graph.

An inflection point is where the graph of a function changes its concavity. This means the curve switches from curving upwards to downwards or vice versa. Inflection points can be found by analyzing where the second derivative is zero or undefined, and where its sign changes.For our function, we compute: \[ f''(x) = -\frac{16}{9}x^{-5/3} + \frac{4}{9}x^{-2/3} \].When checking for inflection points, we find \( f''(x) \) does not equal zero at \( x = 0 \), and the function's second derivative is undefined at this point. However, there's a critical change in the sign of the second derivative at \( x = 0 \).The transition from positive concavity when \( x < 0 \) to negative concavity for \( x > 0 \) signals an inflection point occurs.Thus, even though \( x=0 \) is a vertical asymptote in terms of \( f''(x) \), it marks a change in concavity for the original function.

The first derivative of a function, \( f'(x) \), tells us about the slope of the function at any given point. It is critical for finding where the function has peaks, valleys, or flat spots, known as local extrema. Calculating the first derivative involves differentiating function \( f(x) \). In our situation, this results in: \[ f'(x) = \frac{8}{3}x^{-2/3} + \frac{4}{3}x^{1/3} \]. We set \( f'(x) = 0 \) to identify critical points. These points are where the slope of the function is zero, so they could be locations for local maxima, minima, or points of inflection. By solving \( \frac{8}{3}x^{-2/3} + \frac{4}{3}x^{1/3} = 0 \), we find\( x = -2 \). Once we have these critical points, further tests, like the second derivative test, are used to determine the exact nature of these points. This step is foundational for conducting further analyses related to the behavior of the function.