Problem 15 - Applied Calculus

Question

Use the first derivative to find all critical points and use the second derivative to find all inflection points. Use a graph to identify each critical point as a local maximum, a local minimum, or neither. \(f(x)=x^{4}-2 x^{2}\)

Step-by-Step Solution

Answer

Critical points at \(x=0\) (local max), \(x=1\), \(x=-1\) (local minima); inflection points at \(x=\pm\frac{1}{\sqrt{3}}\).

Step 1: Find the First Derivative

To find the critical points, we first need the first derivative of the function. The given function is \(f(x)=x^4-2x^2\). The first derivative \(f'(x)\) will be obtained by differentiating each term separately.\[f'(x)=4x^3-4x\]

Step 2: Find Critical Points

Critical points occur where the first derivative is equal to zero or is undefined. For the function \(f(x)=x^4-2x^2\), the first derivative was found to be \(f'(x)=4x^3-4x\). Let's set this equal to zero and solve for \(x\):\[4x(x^2-1)=0\]This equation factors further:\[4x(x-1)(x+1)=0\]Setting each factor to zero gives us the critical points: \(x=0\), \(x=1\), and \(x=-1\).

Step 3: Find the Second Derivative

Next, we find the second derivative to help determine inflection points and confirm the nature of each critical point. The first derivative is \(f'(x) = 4x^3-4x\). Differentiate again:\[f''(x) = 12x^2 - 4\]

Step 4: Identify Inflection Points

Inflection points occur where the second derivative changes sign. Set \(f''(x)\) equal to zero to find potential inflection points:\[12x^2-4=0\]Solving this, we find:\[12x^2=4\]\[x^2=\frac{1}{3}\]\[x=\pm\frac{1}{\sqrt{3}}\]

Step 5: Use Second Derivative Test for Critical Points

To classify each critical point, use the second derivative test. Substitute each critical point into \(f''(x) = 12x^2 - 4\):- For \(x=0\): \(f''(0) = -4\) (negative, so \(x=0\) is a local maximum)- For \(x=1\): \(f''(1) = 8\) (positive, so \(x=1\) is a local minimum)- For \(x=-1\): \(f''(-1) = 8\) (positive, so \(x=-1\) is a local minimum)

Step 6: Confirm with Graph Examination

Graph the function \(f(x)=x^4-2x^2\) to visually identify the nature of each critical point. At \(x=0\), the graph peaks, confirming a local maximum. At \(x=1\) and \(x=-1\), the graph has troughs, confirming local minima.

Key Concepts

First Derivative

The first derivative of a function, denoted as \(f'(x)\), is an essential tool in calculus. It helps us understand how the function is changing at any given point. Specifically, the first derivative gives us the slope of the tangent line to the curve at each point.
For the function \(f(x)=x^4-2x^2\), the first derivative is found by differentiating each term separately:

• The derivative of \(x^4\) is \(4x^3\).
• The derivative of \(-2x^2\) is \(-4x\).

So, the first derivative is \(f'(x) = 4x^3 - 4x\).
The critical points of a function are where this derivative is equal to zero or undefined. In our case, we look for solutions to the equation \(4x(x^2-1)=0\), resulting in the critical points \(x=0\), \(x=1\), and \(x=-1\).
Identifying these points is crucial because they might be areas where the function reaches local maxima or minima.

Second Derivative

The second derivative of a function, denoted as \(f''(x)\), provides insight into the function's concavity. Concavity refers to whether the function is curving upwards or downwards at a particular point. This can help us determine the nature of critical points and find inflection points.

For the function \(f(x)=x^4- 2x^2\), we have already found the first derivative \(f'(x) = 4x^3 - 4x\). Differentiating again, we obtain the second derivative:

• \(f''(x) = 12x^2 - 4\).

The second derivative allows us to perform the second derivative test, aiding in classifying critical points as local maxima or minima, and to locate inflection points where the concavity changes.

Local Maximum

A local maximum occurs at a point where the function has a higher value than at any nearby points. To classify a critical point as a local maximum using derivatives, we use the second derivative test.
For the given function, the critical point \(x=0\) was tested with the second derivative:

• \(f''(0) = -4\), which is less than 0.

A negative second derivative at a critical point indicates that the graph is concave down at that point, confirming a local maximum. Thus, at \(x=0\), the function peaks, meaning it reaches a local maximum.

Local Minimum

A local minimum is a point where the function takes a smaller value than at any nearby points. To identify a local minimum, the second derivative test can be used effectively.
For the function \(f(x)=x^4-2x^2\), critical points \(x=1\) and \(x=-1\) are evaluated:

• For \(x=1\), \(f''(1) = 8\), which is greater than 0, indicating concave up, thus a local minimum.
• Similarly, for \(x=-1\), \(f''(-1) = 8\), also greater than 0, confirming another local minimum.

Therefore, the function forms a U-shape at these points, suggesting these are the low troughs on the graph indicating local minima.

Inflection Points

Inflection points are points on a curve where the curvature changes direction. At an inflection point, the graph of a function changes from concave up to concave down or vice versa.
To find potential inflection points for the function \(f(x) = x^4 - 2x^2\), we set the second derivative equal to zero.

• \(f''(x) = 12x^2 - 4 = 0\)
• Solving this gives \(x=\pm \frac{1}{\sqrt{3}}\).

These points are where the second derivative changes sign, indicating transitions between different curvature types. They help create a more rounded understanding of the graph's behavior in these regions.