Question

In Exercises \(11-18,\) use analytic methods to find the extreme values of the function on the interval and where they occur. $$f(x)=x^{3 / 5}, \quad-2< x \leq 3$$

Short answer

The function \(f(x) = x^{3 / 5}\) reaches an absolute minimum at \(x = 0\) and an absolute maximum at \(x = 3\).

Step-by-step solution

Find the Derivative

First, calculate the derivative of the function \(f(x) = x^{3 / 5}\). This yields \(f'(x) = \frac{3}{5} x^{-2 / 5}\).

Find the Critical Points

Set the derivative equal to zero to find the critical points: \(\frac{3}{5} x^{-2 / 5} = 0\). This gives \(x = 0\) as the only critical point.

Evaluate the Function at the Endpoints and Critical Point

Evaluate the function at the endpoints (-2 and 3) and at the critical point (0). For \(x = -2\), \(f(-2) = (-2)^{3 / 5}\). For \(x = 3\), \(f(3) = 3^{3 / 5}\). And, for \(x = 0, f(0) = 0\).

Identify the Maximum and Minimum Values

Identify the maximum and minimum values of the function by comparing the values from Step 3. The function will have its absolute minimum at \(x = 0\), and its maximum at \(x = 3\).

Key concepts

Analytic Methods in Calculus

When it comes to analyzing functions and finding their extreme values, calculus provides a systematic approach: analytic methods. This approach often involves taking the derivative of the function, which is the cornerstone of finding where the function's slope is zero—these points typically indicate a local maximum, minimum, or saddle point.

In the context of the given exercise, the function under consideration is a power function. Analytic methods allow us to not only find the derivative but also to evaluate the behavior of the function at boundary points of the given interval. After all, extreme values can also occur at the endpoints of the interval, not just where the derivative is zero. By carefully analyzing both the critical points and the endpoints, we can identify the function's extreme values within the specified range.

Finding Critical Points

Critical points are pivotal in determining the highs and lows of a function's graph. To find these points, we first need to find the derivative of the function. A critical point occurs where the derivative is zero or undefined.

In our exercise, we established that the only critical point of the function is at x = 0, since it's the only solution to the equation \(\frac{3}{5} x^{-2 / 5} = 0\). Note that the function does not have a critical point where the derivative is undefined within the given interval. After identifying the critical point, we must compare the value of the function at this point to the values at the endpoints to determine if it's an extreme value. It's valuable for students to remember that not all critical points yield an extreme value; some may be inflection points, where the function changes concavity but doesn't achieve a maximum or minimum.

Derivative of a Function

The derivative of a function is fundamental in calculus, representing the rate at which the function's value changes. It can show us the slope of the function at any given point and is thus instrumental in finding the function's critical points and potential extrema.

In our example, the derivative of \(f(x) = x^{3/5}\) is \(f'(x) = \frac{3}{5} x^{-2/5}\), which simplifies our task of finding critical points. When students understand that the derivative is the key to finding when and where a function levels out or changes direction, they'll be better equipped to tackle more complex problems involving rates of change and optimizing functions. Remember, a non-zero derivative means the function is increasing or decreasing at that point, while a zero derivative implies the possibility of an extreme value.