Problem 26 - Calculus of a Single Variable

Question

Applying the First Derivative Test In Exercises \(17-40\) , (a) find the critical numbers of \(f\) (if any), (b) find the open interval(s) on which the function is increasing or decreasing, (c) apply the First Derivative Test to identify all relative extrema, and (d) use a graphing utility to confirm your results. $$ f(x)=x^{4}-32 x+4 $$

Step-by-Step Solution

Answer

The critical number is \(x = 2\). The function is decreasing on the interval \((-\infty, 2)\) and increasing on the interval \((2, \infty)\). There is a relative minimum at \(x = 2\).

Step 1: Find the Critical Numbers

To find the critical numbers, first the derivative of the function must be calculated. The derivative of \(f\) is \(f'(x)=4x^{3}-32\). Setting this equal to zero gives the critical numbers when solved for \(x\). Hence \(f'(x)=0\) implies \(x^{3}=8\), thus the critical numbers are \(x=2\).

Step 2: Find the Intervals

To find the increasing and decreasing intervals, the sign of the derivative should be considered. The derivative \(f'(x) > 0\) when \(x > 2\) and \(f'(x) < 0\) when \(x < 2\). This means that the function is increasing when \(x > 2\) and decreasing when \(x < 2\).

Step 3: Apply the First Derivative Test

Applying the First Derivative Test involves looking at the sign of the derivative before and after the critical numbers. Since \(f'(x) < 0\) when \(x < 2\) and \(f'(x) > 0\) when \(x > 2\), it is concluded that \(x = 2\) is a relative minimum.

Step 4: Confirm the Solutions

The critical number, increasing and decreasing intervals and the relative minimum can be validated by using a graphing utility to plot the function and its derivative. Upon doing this, it will be observed that the function has a minimum at \(x = 2\), decreases for \(x < 2\) and increases for \(x > 2\), confirming the above findings.

Key Concepts

Critical Numbers

Critical numbers play a crucial role in understanding a function's behavior. They are the points where the derivative of a function is either zero or undefined. These numbers help pinpoint potential turning points or points where the function changes direction.

For the function given, \( f(x) = x^4 - 32x + 4 \), its first derivative is \( f'(x) = 4x^3 - 32 \). To find critical numbers, we set \( f'(x) = 0 \). Solving \( 4x^3 - 32 = 0 \) gives \( x^3 = 8 \), leading to \( x = 2 \).

This tells us that \( x = 2 \) is a critical number and suggests that around this point, the function might change from increasing to decreasing or vice versa.

Increasing Intervals

Identifying where a function is increasing is essential for understanding its overall behavior. A function is increasing when its derivative is positive. This positivity implies the slope is upward as you move along the curve.

For the function \( f(x) = x^4 - 32x + 4 \), we've established that its derivative is \( f'(x) = 4x^3 - 32 \). To determine the increasing intervals, look at where \( f'(x) > 0 \).

Solving this inequality for \( f'(x) \), we find that for \( x > 2 \), \( f'(x) \) is positive. Hence, the function is increasing on the interval \( (2, \infty) \). This means as \( x \) grows beyond 2, the value of \( f(x) \) keeps rising.

Decreasing Intervals

Understanding decreasing intervals of a function gives insight into where the function values are dropping. A function decreases in intervals where its derivative is negative, meaning the slope of the tangent line to the graph is downward.

In our example, with \( f(x) = x^4 - 32x + 4 \), the derivative \( f'(x) = 4x^3 - 32 \) helps determine these intervals. By analyzing \( f'(x) < 0 \), we can identify where the function descends.

When evaluating \( f'(x) \), we discover that for \( x < 2 \), \( f'(x) \) is negative. Therefore, the function is decreasing on the interval \( (-\infty, 2) \). This indicates that as \( x \) approaches 2 from the left, \( f(x) \) decreases.

Relative Extrema

A relative extremum is a point where the function changes from increasing to decreasing or vice versa, resulting in a relative maximum or minimum. The First Derivative Test provides a systematic approach to find these points.

By using the First Derivative Test on our function \( f(x) = x^4 - 32x + 4 \), we examine the change in the sign of the derivative \( f'(x) = 4x^3 - 32 \) around the critical number \( x = 2 \).

Before \( x = 2 \), \( f'(x) \) is negative, and after \( x = 2 \), \( f'(x) \) becomes positive. This sign change indicates a transition from decreasing to increasing, showing a relative minimum at \( x = 2 \). Hence, the function reaches a low point at this critical number, which reveals a critical characteristic of its curve.