Problem 18 - Single Variable Calculus

Question

(a) find the intervals on which \(f\) is increasing or decreasing, and (b) find the relative maxima and relative minima of \(\vec{f}\). $$ f(x)=x^{1 / 3}-x^{2 / 3} $$

Step-by-Step Solution

Answer

The function \(f(x) = x^{1/3} - x^{2/3}\) is increasing on the interval \(-\infty < x < 0\), decreasing on the interval \(0 < x < \infty\), and has a relative maximum at x = 0. There are no relative minima for this function.

Step 1: Find the derivative of f(x)

To find the derivative of the given function, $$ f(x) = x^{1/3} - x^{2/3}, $$ we need to use the power rule: $$ \frac{d}{dx} x^n = nx^{n-1} $$ Using this rule on the given function, we get the derivative as: $$ f'(x) = \frac{1}{3}x^{-2/3} - \frac{2}{3}x^{-1/3} $$

Step 2: Find critical points

To find critical points, set f'(x) equal to zero and solve for x: $$ 0 = \frac{1}{3}x^{-2/3} - \frac{2}{3}x^{-1/3} $$ Combine the terms with common denominators: $$ 0 = x^{-1/3}(\frac{1}{3}x^{-1} - \frac{2}{3}) $$ Now, check for x-values that make either factor equal to zero. The second factor is never zero, but the first factor x^{-1/3} is zero when x = 0. Therefore, x = 0 is a critical point.

Step 3: Determine the intervals of increase or decrease

We'll analyze the signs of f'(x) in the intervals created by the critical point x = 0: - For \(x < 0\), both factors of f'(x) are negative, so f'(x) is positive. Therefore, the function is increasing on the interval \(-\infty < x < 0\). - For \(x > 0\), the first factor of f'(x) is positive, and the second factor is negative, so f'(x) is negative. Therefore, the function is decreasing on the interval \(0 < x < \infty\).

Step 4: Identify relative maxima and minima

Based on the intervals of increase and decrease, we can determine the relative maxima and minima of the function: - Since the function is increasing as x approaches 0 from the left, and then decreasing as x moves away from 0 to the right, there is a relative maximum at the critical point x = 0. - There are no relative minima for this function. In conclusion, the function is increasing on the interval \(-\infty < x < 0\), decreasing on the interval \(0 < x < \infty\), and has a relative maximum at x = 0. There are no relative minima for this function.

Key Concepts

Derivatives

Derivatives are a cornerstone of calculus, representing how a function changes as its input changes. To find the derivative of a function means to determine this rate of change. In this exercise, we apply the power rule, which is:

• The derivative of \( x^n \) is \( nx^{n-1} \).

For the function \( f(x) = x^{1/3} - x^{2/3} \), applying the power rule gives: \[ f'(x) = \frac{1}{3}x^{-2/3} - \frac{2}{3}x^{-1/3} \]This derivative tells us how the function's value is changing with respect to \( x \), and it's essential for finding critical points and understanding the behavior of the function on different intervals.
By understanding derivatives, you gain insights into the function's slope and direction of change at any point.

Increasing and Decreasing Intervals

Once the derivative is found, it tells us where the function is increasing or decreasing. If the derivative (\( f'(x) \)) is positive on an interval, the function is increasing there, meaning it slopes upwards. If \( f'(x) \) is negative, the function is decreasing, or sloping downwards.

• For \( f(x) = x^{1/3} - x^{2/3} \), we calculated \( f'(x) = \frac{1}{3}x^{-2/3} - \frac{2}{3}x^{-1/3} \).

After identifying the critical points (where \( f'(x) = 0 \) or undefined), we determine the sign of \( f'(x) \).

• This function is increasing from \( -\infty \) to \( 0 \), as the signs recorded around these points indicate.
• It starts decreasing at \( x = 0 \) to \( +\infty \).

Understanding these intervals is crucial for graphing the function and predicting where it will peak or dip.

Critical Points

Critical points are where the function's derivative equals zero or is undefined, indicating potential peaks, troughs, or flat regions. These are key for analyzing a function's graph because they show where the function transitions between increasing and decreasing.

• To find critical points, solve \( f'(x) = 0 \). For our function:

\[\frac{1}{3}x^{-2/3} - \frac{2}{3}x^{-1/3} = 0\]Solving this gives \( x = 0 \) as a critical point. These points must be checked because they often indicate relative maxima or minima. Knowing how to spot these points helps in understanding the overall trend of a function.

Relative Maxima and Minima

Relative maxima and minima occur at critical points where a function changes direction. A relative maximum is a point where the function reaches a peak locally, before decreasing afterward. A relative minimum is a local trough where the function increases afterward.

• We determined that for \( f(x) = x^{1/3} - x^{2/3} \), there is a relative maximum at \( x = 0 \).

Before reaching \( x = 0 \), the function is increasing, and after passing \( x = 0 \), it begins decreasing. This pattern defines a maximum at this point, though it’s only relative to its immediate surroundings.

• It's important to note there is no relative minimum according to the behavior on either side of \( x = 0 \).

Understanding these helps with sketching the graph of the function and predicting its behavior on various intervals.