Question

In Exercises, find the critical numbers and the open intervals on which the function is increasing or decreasing. Then use a graphing utility to graph the function. $$ h(x)=x \sqrt[3]{x-1} $$

Short answer

There are no real critical numbers. The function is decreasing over the entire domain.

Step-by-step solution

Find the Derivative

The original function \(h(x) = x \sqrt[3]{x-1}\) can be rewritten in this more convenient form \(h(x) = x (x-1)^{1/3}\). To find its derivative, use the product rule for differentiation, which states that the derivative of a product of two functions is the first function times the derivative of the second function plus the second function times the derivative of the first function. After applying the product rule, obtain the derivative, \(h'(x) = (x-1)^{1/3} + x (1/3)(x-1)^{-2/3}\).

Find Critical Numbers

Critical numbers are points where the derivative is either equal to zero or does not exist. So, we set our derivative equal to zero: \(h'(x) = 0\), which simplifies to \((x-1)^{1/3} + x (1/3)(x-1)^{-2/3} = 0\). From this, we find there are no real solutions for x, and hence there are no critical numbers.

Establish Intervals

In this case, without any critical numbers, we only have one interval to consider \(-(∞, ∞)\). Choose a test point, such as x = 0, and substitutes it into the derivative. The sign of the derivative at this point will tell whether the function is increasing or decreasing in this interval. Calculating, we get \(h'(0) = -1/3 < 0\), which means the function is decreasing over the interval \(-(∞, ∞)\).

Key concepts

Derivative

In calculus, a derivative represents the rate at which a function changes at any given point. It provides us the slope of a function at a particular point and is fundamental to analyzing functions. In our given function \(h(x) = x \sqrt[3]{x-1}\), when rewritten as \(h(x) = x (x-1)^{1/3}\), the derivative helps us determine where the function is increasing or decreasing, as well as identifying critical points.
To find the derivative of our function, we used the product rule. This rule is essential when differentiating products of two simpler functions. By correctly applying it, we arrived at the derivative: \(h'(x) = (x-1)^{1/3} + x \left(\frac{1}{3}(x-1)^{-2/3}\right)\). This form allows us to analyze the behavior of \(h(x)\) across different intervals.

Product Rule

The product rule is a derivative rule used when dealing with products of two or more functions. Simply put, if you have a function that is a product of two functions, such as \(u(x)\) and \(v(x)\), their derivative is calculated as: \((uv)' = u'v + uv'\). This rule simplifies the process of differentiation for complex functions.
In the exercise, the function \(h(x) = x (x-1)^{1/3}\) involves a product between \(x\) and \((x-1)^{1/3}\). By applying the product rule:

• Calculate the derivative of \(u(x)=x\), which is \(1\).
• Calculate the derivative of \(v(x)=(x-1)^{1/3}\). Use the power rule to get \(\frac{1}{3}(x-1)^{-2/3}\).
• Combine them using the formula to derive \(h'(x)\).

This step is crucial for finding how the function behaves and for determining the critical numbers.

Increasing and Decreasing Intervals

To identify increasing and decreasing intervals, we use the derivative. A positive derivative indicates that a function is increasing, while a negative derivative suggests it is decreasing within that interval. Generally, we also check where the derivative equals zero or is undefined to find critical points that help define these intervals.
For \(h(x)\), the derivative \(h'(x)\) had no real solutions equal to zero, meaning no critical numbers were found where it changes from increasing to decreasing or vice versa. Hence, the function is analyzed over the interval \((−∞, ∞)\). Evaluating the derivative at a test point, such as \(x = 0\), showed \(h'(0) = -1/3\), indicating the function is consistently decreasing across this entire interval.

Graphing Utility

Graphing utilities are powerful tools that visually represent mathematical equations, which helps in better understanding their behavior. These tools can plot the graphs instantly once you enter the function. Seeing the graphical representation of \(h(x) = x \sqrt[3]{x-1}\) allows students to confirm their findings from calculus visually.
When this function is graphed, you'll notice the consistent downward slope, corroborating that \(h(x)\) is decreasing across the interval \((−∞, ∞)\), as indicated by our derivative analysis. Using such utilities not only aids exacting calculations but also serves as a visual check for calculated intervals and critical points.