Problem 30 - Essential Calculus

Question

Find the critical numbers of \(f\) (if any). Find the open intervals on which the function is increasing or decreasing and locate all relative extrema. Use a graphing utility to confirm your results. $$ f(x)=(x-1)^{1 / 3} $$

Step-by-Step Solution

Answer

The function \(f(x)=(x-1)^{1 / 3}\) has one critical number at \(x=1\). It is increasing over the entire real number line. There are no relative extrema.

Step 1: Find the derivative of the function

The derivative of \(f(x)=(x-1)^{1 / 3}\) can be found using the chain rule. The derivative of \(f(x)\) is given by \(f'(x)=\frac{1}{3}(x-1)^{-2/3}\).

Step 2: Find the critical numbers

By setting the derivative to zero and undefined, we can find the critical numbers. \(f'(x)=0\) has no solution, and \(f'(x)\) is undefined at \(x = 1\), so the critical number is \(x = 1\).

Step 3: Determine intervals of increase and decrease

With \(x < 1\), \(f'(x) > 0\) meaning the function is increasing, and with \(x > 1\), \(f'(x) > 0\) meaning the function is still increasing. So, the function is increasing for all real numbers.

Step 4: Identify relative extrema

Since the function is always increasing, there are no relative maximum or minimum points.

Step 5: Verify results with graphing utility

The steps taken in this solution can be verified using a graphing utility, such as a graphing calculator or software like Desmos. The graph of \(f(x)\) should appear to be always increasing, with a critical point at \(x = 1\).

Key Concepts

Derivative

The derivative is a fundamental concept in calculus that represents the rate of change of a function. It is essentially finding how a function behaves at different points. For the function given, \(f(x) = (x-1)^{1/3}\), the derivative \(f'(x)\) can be determined using the chain rule, a vital differentiation technique. Here, the function can be thought of as a composition where the outer function is a cube root and the inner function is \(x - 1\). Using the chain rule, we find:

• The derivative of the outer function: \(\frac{1}{3}(u)^{-2/3}\).
• The derivative of the inner function: \(1\).

Therefore, applying these, the derivative of \(f(x)\) is \(f'(x) = \frac{1}{3}(x-1)^{-2/3}\). This indicates how steep or flat the tangent is at any point \(x\). By calculating this derivative, we obtain important insights into the behavior of the function at each point.

Intervals of Increase/Decrease

To understand where a function is increasing or decreasing, we analyze the sign of its derivative. If the derivative \(f'(x) > 0\) on an interval, then the function is increasing on that interval. Conversely, if \(f'(x) < 0\), the function is decreasing. For the given function, the derivative is \(f'(x) = \frac{1}{3}(x-1)^{-2/3}\), which never becomes negative. Breaking it down:

• For \(x < 1\), \(f'(x)\) is positive, indicating that the function is increasing.
• For \(x > 1\), \(f'(x)\) remains positive, showing that the function continues to increase.

It turns out that the function is indeed increasing for all \(x eq 1\). Since \(f'(x)\) is undefined at \(x = 1\), we must reconsider such points separately. A graphing utility can further help confirm these findings visually by plotting the function, showing a constant upward trend.

Relative Extrema

Relative extrema refer to points in the domain of a function where the function reaches a local minimum or maximum value. Identifying these helps in understanding the function's turning points. To find relative extrema, we need critical numbers, where the derivative of the function is zero or undefined.In this case, after determining \(f'(x)\), we see:

• \(f'(x) = 0\) has no solutions, implying no horizontal tangents or apparent peaks or valleys.
• \(f'(x)\) is undefined at \(x = 1\), marking a potential critical point.

Despite this critical point, since \(f(x)\) is always increasing and does not change direction, it does not possess any relative minima or maxima. Therefore, there are no local turning points for this function. The critical point at \(x = 1\) does not signify a change in direction but confirms the nature of the critical point as neither a maximum nor a minimum.