Question

A function \(f(x)\) is given. (a) Give the domain of \(f\). (b) Find the critical numbers of \(f\). (c) Create a number line to determine the intervals on which \(f\) is increasing and decreasing. (d) Use the First Derivative Test to determine whether each critical point is a relative maximum, minimum, or neither. \(f(x)=\frac{(x-2)^{2 / 3}}{x}\)

Short answer

Domain: \((- fty, 0)\cup(0, fty)\). Critical number: \(x=2\). Decreasing on \((0,2)\), increasing on \((- fty,0)\) and \((2,\infty)\). Relative minimum at \(x=2\).

Step-by-step solution

Determine the domain of \( f \)

The domain of a function consists of all possible input values \( x \) for which the function is defined. For \( f(x) = \frac{(x-2)^{2/3}}{x} \), the function is undefined when the denominator is zero or when the expression under the root sign leads to a negative value for an even root. In this case, since \( x \) is in the denominator, \( x = 0 \) is not allowed. Therefore, the domain of \( f \) is all real numbers except \( x = 0 \), expressed as \( (-\infty, 0) \cup (0, \infty) \).

Find the critical numbers of \( f \)

A critical number is a number \( c \) in the domain of \( f \) where the derivative \( f'(x) \) is zero or undefined. First, find \( f'(x) \). To do this, apply the quotient rule \( \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \).For \( u = (x-2)^{2/3} \) and \( v = x \), we have:- \( u' = \frac{2}{3}(x-2)^{-1/3} \)- \( v' = 1 \)The derivative is:\[ f'(x) = \frac{\frac{2}{3}(x-2)^{-1/3} \cdot x - (x-2)^{2/3} \cdot 1}{x^2} \]Simplify to find:\[ f'(x) = \frac{2}{3}x(x-2)^{-1/3} - (x-2)^{2/3} \over x^2 \]Set \( f'(x) = 0 \) and \( f'(x) \) is undefined for critical points:Where \( x = 2 \) makes the numerator zero. The derivative is also undefined at \( x = 0 \), but \( x = 0 \) is not in the domain.Thus, the critical number is \( x = 2 \).

Determine intervals of increase and decrease

Analyze the sign of \( f'(x) \) around the critical point \( x = 2 \). Check the sign of \( f'(x) \) within intervals created by the critical point and undefined points:1. Pick a test point for \( (-\infty, 0) \); choose \( x = -1 \).2. Pick a test point in \( (0, 2) \); choose \( x = 1 \).3. Pick a test point in \( (2, \infty) \); choose \( x = 3 \).Calculate \( f'(x) \) at these points:- For \( x = -1 \), \( f'(x) > 0 \).- For \( x = 1 \), \( f'(x) < 0 \).- For \( x = 3 \), \( f'(x) > 0 \).Thus, \( f \) is increasing on \((-\infty, 0)\) and \((2, \infty)\), and decreasing on \((0, 2)\).

Use the First Derivative Test

Apply the First Derivative Test to determine the nature of the critical point at \( x = 2 \). From Step 3, we know:- The function changes from decreasing to increasing at \( x = 2 \).Therefore, according to the First Derivative Test, \( x = 2 \) is a relative minimum.

Key concepts

Domain of a Function

In calculus, the domain of a function is crucial because it tells us the set of all possible input values (usually represented as \(x\)) that the function can accept. For example, the function \( f(x) = \frac{(x-2)^{2/3}}{x} \) has specific restrictions on its domain. This is because any value that makes the denominator zero or results in an undefined root causes the function to become undefined.

For this particular function, the denominator is simply \(x\). This means \(x\) cannot be zero, because division by zero is undefined. Therefore, the domain does not include zero and is expressed as all real numbers except zero. This can be written as the union of two intervals: \((-\infty, 0) \cup (0, \infty)\). Recognizing the domain helps us understand where the function can be evaluated and where it might have breaks or discontinuities in its graph.

Critical Numbers

Critical numbers are points in the domain of a function where its derivative is either zero or undefined, which can indicate potential maxima, minima, or points of inflection.

To find the critical numbers for \( f(x) \), we calculate the derivative using the quotient rule. With \( u = (x-2)^{2/3} \) and \( v = x \), the derivative \( f'(x) \) is complicated but necessary to explore these critical points. After simplifying using the quotient rule, the function is set to zero to solve for critical points where the derivative equals zero. This leads us to \(x = 2\). Although \( f'(x) \) is also undefined at \( x = 0 \), \( x = 0 \) is not within the function's domain. Thus, \( x = 2 \) is our only critical number for further analysis.

First Derivative Test

The First Derivative Test helps us determine the nature of critical points—whether they are local maxima, minima, or neither. This involves examining the signs of the derivative before and after each critical point.

For \( x = 2 \), identified as a critical number, notice how the derivative \( f'(x) \) transitions from negative to positive values. This sign change indicates that the slope of the tangent to the function changes from decreasing to increasing. According to the First Derivative Test, this reflects that \( x = 2 \) is a point where the function has a relative minimum. This detailed analysis gives us insights into the behavior of the function at particular points within its domain.

Increasing and Decreasing Intervals

Determining the intervals where a function increases or decreases involves finding where the derivative is positive (increasing) or negative (decreasing).

For \( f(x) \), after calculating and examining \( f'(x) \), we pick test points in each interval defined by the critical point \( x = 2 \) and the undefined point \( x = 0 \).

• For \( x = -1 \), \( f'(x) > 0 \), so \( f(x) \) is increasing on \((-\infty, 0)\).
• For \( x = 1 \), \( f'(x) < 0 \), thus the function decreases on \((0, 2)\).
• For \( x = 3 \), \( f'(x) > 0 \), and so it rises on \((2, \infty)\).

This analysis reveals the behavior of the function over its domain. Knowing these intervals is key to understanding how the function behaves over different numbers, depicting a more comprehensive picture of its graph.