Problem 14 - Calculus Single Variable

Question

Use the first derivative to determine the intervals on which the given function \(f\) is increasing and on which \(f\) is decreasing. At each point \(c\) with \(f^{\prime}(c)=0,\) use the First Derivative Test to determine whether \(f(c)\) is a local maximum value, a local minimum value, or neither. $$ f(x)=1 /\left(x^{4}+6\right) $$

Step-by-Step Solution

Answer

Function is increasing on \((-\infty, 0)\) and decreasing on \((0, \infty)\); local maximum at \(x=0\).

Step 1: Find the First Derivative

To analyze where the function is increasing or decreasing, first find the derivative of the function. The function is given by \[ f(x) = \frac{1}{x^4 + 6}. \]Using the quotient rule or rewrite it via the chain rule as a power: \[ f(x) = (x^4 + 6)^{-1}. \]The derivative is \[ f'(x) = -1 \times (x^4 + 6)^{-2} \times 4x^3 = \frac{-4x^3}{(x^4 + 6)^2}. \]

Step 2: Determine Critical Points

Critical points occur where the derivative is zero or undefined. Set \[ f'(x) = \frac{-4x^3}{(x^4 + 6)^2} = 0. \]This happens when the numerator is zero, which is \[ -4x^3 = 0, \]leading to \[ x = 0. \]The derivative is defined for all real numbers, as the denominator \( (x^4 + 6)^2 \) is never zero, since \( x^4 + 6 \geq 6. \) Therefore, the only critical point is at \( x = 0. \)

Step 3: Test Intervals Around Critical Points

The critical point divides the number line into intervals. Test each interval to determine the sign of \( f'(x) \). - For \( x < 0 \), choose \( x = -1 \): \[ f'(-1) = \frac{-4(-1)^3}{((-1)^4 + 6)^2} = \frac{4}{49} > 0. \] Thus, \( f(x) \) is increasing on this interval.- For \( x > 0 \), choose \( x = 1 \): \[ f'(1) = \frac{-4(1)^3}{(1^4 + 6)^2} = \frac{-4}{49} < 0. \] Thus, \( f(x) \) is decreasing on this interval.

Step 4: Apply the First Derivative Test

At \( x = 0 \), \( f(x) \) changes from increasing to decreasing across the point. According to the First Derivative Test, this behavior indicates a local maximum.Thus, \( f(0) \) is a local maximum.

Key Concepts

Critical Points

Critical points of a function occur where the first derivative is zero or undefined. These are the potential spots for local maxima or minima because this is where the slope of the tangent to the function's graph is flat. For the function \[f(x) = \frac{1}{x^4 + 6},\]we found that its derivative is \[f'(x) = \frac{-4x^3}{(x^4 + 6)^2}.\]To find the critical points, set the derivative equal to zero. Here, \[-4x^3 = 0\]leads to the critical point \[x = 0.\]The derivative is defined for all real numbers since the denominator \[(x^4 + 6)^2\] is never zero. Therefore, the only critical point is at \[x = 0.\] This means the tangent line is horizontal at this point, indicating a possible peak or valley on the graph of the function.

Increasing and Decreasing Functions

Knowing where a function increases or decreases is crucial for understanding its graph's overall shape. This is determined by the sign of the first derivative:

• If the derivative \(f'(x) > 0\), the function is increasing. An increasing function climbs as we move from left to right.
• If \(f'(x) < 0\), the function is decreasing. On a graph, this means the curve is moving downward as we proceed from left to right.

For the function given in the exercise, we analyze intervals around the critical point \(x = 0\). We find

• For \(x < 0\), testing \(x = -1\) gives \(f'(-1) = \frac{4}{49} > 0\). Hence, the function is increasing for \(x < 0\).
• For \(x > 0\), testing \(x = 1\) gives \(f'(1) = \frac{-4}{49} < 0\). Therefore, the function decreases for \(x > 0\).

These insights help in sketching the curve and understanding its behavior across different intervals.

Local Maximum and Minimum

Local maxima and minima are points where the function reaches a peak or valley in its immediate vicinity. We can use the First Derivative Test to determine their existence and nature:

• A local maximum occurs at a point \(c\) where the derivative changes sign from positive to negative.
• A local minimum occurs where the derivative changes from negative to positive.

At the critical point \(x = 0\) for the function \[f(x) = \frac{1}{x^4 + 6},\]we observed:

• For \(x < 0\), \(f'(x) > 0\) indicating the function is increasing as we approach from the left.
• For \(x > 0\), \(f'(x) < 0\) showing the function decreases as we move to the right.

The function transitions from increasing to decreasing at \(x = 0\), revealing it to be a local maximum. The First Derivative Test helps confirm this finding as it shows how the function behaves around the critical point.