Problem 125 - Thomas Calculus

Question

Find the inflection points (if any) on the graph of the function and the coordinates of the points on the graph where the function has a local maximum or local minimum value. Then graph the function in a region large enough to show all these points simultaneously. Add to your picture the graphs of the function's first and second derivatives. How are the values at which these graphs intersect the \(x\)-axis related to the graph of the function? In what other ways are the graphs of the derivatives related to the graph of the function? \begin{equation}\begin{array}{l}{\text { Graph } f(x)=2 x^{4}-4 x^{2}+1 \text { and its first two derivatives }} \\ {\text { together. Comment on the behavior of } f \text { in relation to the signs }} \\ {\text { and values of } f^{\prime} \text { and } f^{\prime \prime}.}\end{array}\end{equation}

Step-by-Step Solution

Answer

Local maxima at \((0, 1)\); minima at \((1, -1)\) and \((-1, -1)\); inflection points at \(x = \pm \frac{1}{\sqrt{3}}\).

Step 1: Find the Derivatives

First, calculate the first derivative, \( f'(x) \), and the second derivative, \( f''(x) \), of the function \( f(x) = 2x^4 - 4x^2 + 1 \). For the first derivative:\[ f'(x) = \frac{d}{dx}(2x^4 - 4x^2 + 1) = 8x^3 - 8x. \]For the second derivative:\[ f''(x) = \frac{d}{dx}(8x^3 - 8x) = 24x^2 - 8. \]

Step 2: Determine Critical Points

To find critical points of \( f(x) \), set \( f'(x) = 0 \) and solve for \( x \).\[ 8x^3 - 8x = 0 \]Factor the equation:\[ 8x(x^2 - 1) = 0 \]This gives:\[ x = 0, \quad x = \pm 1. \]These are the potential points for local maxima or minima.

Step 3: Determine Inflection Points

Set \( f''(x) = 0 \) to find inflection points:\[ 24x^2 - 8 = 0 \]Solve for \( x \):\[ 24x^2 = 8 \] \[ x^2 = \frac{1}{3} \] \[ x = \pm\frac{1}{\sqrt{3}}. \]These are possible inflection points.

Step 4: Test for Local Maxima and Minima

Using the second derivative test, evaluate \( f''(x) \) at \( x = 0 \), \( x = 1 \), and \( x = -1 \):- At \( x = 0 \), \( f''(0) = -8 \) (negative), indicating a local maximum.- At \( x = 1 \), \( f''(1) = 16 \) (positive), indicating a local minimum.- At \( x = -1 \), \( f''(-1) = 16 \) (positive), indicating a local minimum.Thus, \( x = 0 \) is a local maximum and \( x = \pm 1 \) are local minima.

Step 5: Evaluate the Function at Critical Points

Calculate \( f(x) \) at each critical point:- \( f(0) = 2(0)^4 - 4(0)^2 + 1 = 1 \), local maximum.- \( f(1) = 2(1)^4 - 4(1)^2 + 1 = -1 \), local minimum.- \( f(-1) = 2(-1)^4 - 4(-1)^2 + 1 = -1 \), local minimum.

Step 6: Confirm Inflection Points

Check changes in concavity of \( f(x) \) around \( x = \pm\frac{1}{\sqrt{3}} \):The signs of \( f''(x) \) change around these values, confirming inflection points exist at \( x = \pm\frac{1}{\sqrt{3}} \).

Step 7: Graphical Analysis

Graph \( f(x) \), \( f'(x) \), and \( f''(x) \) together in a plot.- Where \( f'(x) = 0 \) marks critical points and indicates where the slope of \( f(x) \) is zero (peaks and troughs).- Where \( f''(x) = 0 \) marks points of inflection and indicates changes in concavity in \( f(x) \).

Key Concepts

Local Maxima

A local maximum is a point on a graph where the function reaches a peak compared to the surrounding points. This means the function changes from an increasing trend to a decreasing trend at the local maximum.
When analyzing a function, you can find these points by identifying where the first derivative, denoted as\( f'(x) \), is zero or undefined. However, it’s not just enough to find when \( f'(x) = 0 \). You also need to check the second derivative, \( f''(x) \), to confirm whether the point is a local maximum.
When:\[ f''(x) < 0 \] it indicates that the function is concave down at that critical point. This result signifies a local maximum because the slope changes from positive to negative. For our specific function \( f(x) = 2x^4 - 4x^2 + 1 \), we found a local maximum at \( x = 0 \) because \( f''(0) = -8 \). This negative value indicates a peak at \( x = 0 \).

Local Minima

Local minima are the flip side of local maxima. A local minimum is a point where the function dips to its lowest value in a small region compared to the nearby values. At these points, the graph shifts from decreasing to increasing.To detect a local minimum, you start by finding where \( f'(x) = 0 \). This calculation identifies potential critical points. Then, the role of the second derivative, \( f''(x) \), becomes crucial:

• When \( f''(x) > 0 \), it signifies that the function is concave up, like a bowl opening upwards.
• This concavity indicates the presence of a local minimum because the slope switches from negative to positive at this point.

In the case of our function, the computations show that the values \( x = 1 \) and \( x = -1 \) are local minima because \( f''(1) = 16 \) and \( f''(-1) = 16 \), both positive, suggesting these dips in the graph.

First Derivative

The first derivative \( f'(x) \) of a function serves as a tool to understand the slope or gradient of the function at any given point. Specifically, it tells us how steep the function is and in which direction it is going.

• If \( f'(x) > 0 \), the function is increasing, sloping upwards.
• If \( f'(x) < 0 \), the function is decreasing, sloping downwards.
• If \( f'(x) = 0 \), the function has a flat slope, indicating critical points where local maxima or minima may occur.

For our function \( f(x) = 2x^4 - 4x^2 + 1 \), we calculated \( f'(x) = 8x^3 - 8x \). By setting this derivative equal to zero: \( 8x(x^2 - 1) = 0 \), we solved for the critical points: \( x = 0, 1, -1 \). These calculations help identify potential locations for peaks and troughs on the graph.

Second Derivative

The second derivative \( f''(x) \) is fundamentally about acceleration, telling us how the rate of change itself is changing. It’s used to interpret the concavity of the function, which helps identify inflection points and distinguish between local maxima and minima.

• When \( f''(x) > 0 \), the function is concave up, indicating a potential local minimum.
• When \( f''(x) < 0 \), the function is concave down, suggesting a potential local maximum.
• When \( f''(x) = 0 \), it suggests the possibility of an inflection point where the graph changes concavity.

In our example, the second derivative was calculated as \( f''(x) = 24x^2 - 8 \). Solving \( f''(x) = 0 \) gives \( x = \pm\frac{1}{\sqrt{3}} \), these are possible inflection points. Here the function may transition from being concave upward to concave downward, or vice versa. Thus, the second derivative provides deeper insights into the shape and dynamics of the function's graph.