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Finding the derivative is an important skill in calculus, as it helps us understand how a function changes. For the function \(f(x) = (x-2)^{4/3}\), our goal is to find the first and second derivatives. The first derivative \(f'(x)\) tells us the rate of change or slope of the function. To find the first derivative, we need to use the power rule, which is a handy tool for differentiating functions of the form \(x^n\). This rule tells us that if we have a function \(x^n\), its derivative \(\frac{d}{dx}(x^n) = nx^{n-1}\).

• For \(f(x) = (x-2)^{4/3}\), we apply the power rule and get \(f'(x) = \frac{4}{3}(x-2)^{1/3}\).
• The first derivative \(f'(x)\) gives us a formula to calculate the slope of the function at any point \(x\).

After finding the first derivative, we need to take it a step further for inflection points by finding the second derivative \(f''(x)\). This derivative tells us how the rate of change is itself changing, which is crucial for analyzing the concavity of the function. Again, using the power rule, we apply it to \(f'(x)\), and simplify to get \(f''(x) = \frac{4}{9} (x-2)^{-2/3}\). This second derivative will help us in the next stage to find potential inflection points.

The power rule is a fundamental technique in calculus for differentiating functions. It simplifies the process of finding derivatives when dealing with power terms. Understanding this rule well is essential, as it forms the basis for solving many calculus problems. When we have a function in the form of \(x^n\), the power rule allows us to differentiate it quickly:

• The expression \(nx^{n-1}\) is derived where \(n\) is a constant.
• For example, for \((x-2)^{4/3}\), \(n\) is \(4/3\) and applying the rule gives, \(f'(x) = \frac{4}{3}(x-2)^{1/3}\).

This straightforward calculation eases the process of handling more complex functions. The power rule is not just limited to natural numbers; it works for fractional and negative exponents too, as demonstrated in our example. By applying the power rule again, we find the second derivative is \(f''(x) = \frac{4}{9} (x-2)^{-2/3}\). Knowing how to use this rule efficiently can significantly help in transitions from basic to more advanced calculus topics.

Understanding concavity is a vital part of analyzing functions in calculus. Concavity tells us about the curvature of the graph of a function. In simpler terms, it indicates whether a graph is opening upwards or downwards at any particular point. This information is crucial when determining where inflection points lie, as these points mark where the graph changes its concavity. To analyze concavity, we primarily focus on the second derivative, \(f''(x)\):

• If \(f''(x) > 0\), the function is concave up, which means its graph resembles a U-shape.
• If \(f''(x) < 0\), the function is concave down, presenting an upside-down U-shape.
• Inflection points occur where \(f''(x) = 0\) or \(f''(x)\) is undefined, indicating a potential change in concavity.

In our example, \(f''(x) = \frac{4}{9} (x-2)^{-2/3}\), which is undefined at \(x = 2\). This discontinuity suggests that \(x = 2\) is an inflection point. By substituting \(x = 2\) into the original function, we find the corresponding function value, and hence, the inflection point is at \((2, 0)\). Recognizing and analyzing concavity changes are vital for a complete understanding of the behavior of a function's graph.