Question

Absolute maxima and minima Determine the location and value of the absolute extreme values of \(f\) on the given interval, if they exist. $$f(x)=3 x^{2 / 3}-x \text { on }[0,27]$$

Short answer

Answer: The absolute maximum value is 4 at x = 8, and the absolute minimum value is 0 at x = 0 and x = 27.

Step-by-step solution

Find the critical points

To find the critical points, take the first derivative of the given function and set it equal to zero: $$f'(x) = \frac{d}{dx} (3x^{2/3} - x)$$ Applying the power rule for derivatives, $$f'(x) = 2x^{-1/3} - 1$$ Now, set \(f'(x)\) equal to zero to find the critical points: $$2x^{-1/3} - 1 = 0$$ Solve this equation for x: $$x^{-1/3} = \frac{1}{2}$$ Raise both sides of this equation to the power of \(-3\) to isolate x: $$x = 8$$ The critical point of the function \(f(x)\) is at \(x = 8\).

Evaluate the endpoints and critical points

Evaluate the function at the critical point and the endpoints of the interval: 1. \(f(0) = 3(0)^{2/3} - 0 = 0\) 2. \(f(8) = 3(8)^{2/3} - 8 = 12 - 8 = 4\) 3. \(f(27) = 3(27)^{2/3} - 27 = 27 - 27 = 0\) The function has the absolute maximum value at \(x = 8\), with \(f(8) = 4\). The function has an absolute minimum value at both \(x = 0\) and \(x = 27\), with \(f(0) = f(27) = 0\).

Key concepts

Critical Points

Critical points are crucial when it comes to finding absolute extrema of a function on a particular interval. Essentially, a critical point occurs at a specific value of \(x\) where the derivative of the function is either zero or undefined. To determine these points, you'll first need to compute the derivative of the function.

• Find the derivative \(f'(x)\) using differentiation rules.
• Set \(f'(x) = 0\) to find potential critical points.
• Solve this equation to determine the values of \(x\).

In our case, we derived \(f'(x) = 2x^{-1/3} - 1\) and set it equal to zero, which solved to \(x = 8\). This value indicates a critical point for the function \(f(x) = 3x^{2/3} - x\) on the interval \([0, 27]\). Remember that evaluating derivative results must always fall within the original interval for them to be valid critical points.

Derivative

The derivative is a fundamental concept in calculus, representing the rate at which a function is changing at any given point. For this exercise, the derivative helps us understand where the function could potentially switch from increasing to decreasing, or vice versa—key for locating maxima and minima.

• Use the power rule and other derivative rules to find the derivative of a function.
• A derivative \(f'(x)\) set to zero locates critical points which might host an absolute extrema.

In the example provided, the function \(f(x)=3x^{2/3}-x\) is differentiated correctly to yield \(f'(x) = 2x^{-1/3} - 1\). This derivative was essential not just for identifying a critical point in the interval, but also for understanding the behavior of the function across the interval.

Endpoint Evaluation

When examining an interval, it is as important to evaluate the endpoints as finding critical points for determining absolute extrema. The endpoints represent the beginning and the end of an interval—values where the function may also reach its maximum or minimum.

• Identify the interval endpoints \(a\) and \(b\).
• Calculate the function's value at these points.
• Compare the output of these evaluations with the values at the critical points.

In the interval \([0, 27]\), we evaluated \(f(0) = 0\) and \(f(27) = 0\). These computations, along with \(f(8) = 4\), helped us identify that the function \(f(x)\) had its absolute maximum at \(x = 8\) and absolute minimum at the endpoints. Endpoint evaluation ensures that you don't miss out on any potential extrema that may occur at the very boundaries of your interval.