Problem 2 - Thomas Calculus in SI Units

Question

Identify the inflection points and local maxima and minima of the functions graphed. Identify the intervals on which the functions are concave up and concave down. $$y=\frac{x^{4}}{4}-2 x^{2}+4$$

Step-by-Step Solution

Answer

Local minima at \( x = -2, 2 \), maximum at \( x = 0 \); concave up for \( x < -1.15 \) and \( x > 1.15 \), concave down for \( -1.15 < x < 1.15 \).

Step 1: Find the First Derivative

To identify critical points, first compute the derivative of the function. The given function is \( y = \frac{x^4}{4} - 2x^2 + 4 \). The derivative is: \[ y' = x^3 - 4x. \] We'll use this to determine critical points, which are potential local maxima, minima, or points of inflection.

Step 2: Solve for Critical Points

Set the derivative equal to zero to find critical points: \[ x^3 - 4x = 0. \]This simplifies to \[ x(x^2 - 4) = 0 \], giving solutions \( x = 0, x = -2, \text{ and } x = 2. \)

Step 3: Find the Second Derivative

To identify concavity and inflection points, compute the second derivative: \[ y'' = 3x^2 - 4. \] We'll use this to analyze concavity by testing points in each interval divided by the roots found.

Step 4: Determine Concavity and Inflection Points

Find where the second derivative is zero to locate potential inflection points: \[ 3x^2 - 4 = 0 \]. This simplifies to \( x^2 = \frac{4}{3} \), leading to potential inflection points at \( x = \pm \sqrt{\frac{4}{3}} \approx \pm 1.15 \). Next, test intervals around these points to determine concavity.- For \( x < -\sqrt{\frac{4}{3}} \), test \( x = -2 \). \( y''(-2) = 8 \), so it's concave up.- For \( -\sqrt{\frac{4}{3}} < x < \sqrt{\frac{4}{3}} \), test \( x = 0 \). \( y''(0) = -4 \), so it's concave down.- For \( x > \sqrt{\frac{4}{3}} \), test \( x = 2 \). \( y''(2) = 8 \), so it's concave up.

Step 5: Evaluate Function Values at Critical Points

Determine whether critical points are local maxima or minima by substituting into the second derivative, or simply evaluating function values if needed.- At \( x = -2 \) and \( x = 2 \), since \( y'' > 0 \) both are local minima.- At \( x = 0 \), since \( y'' < 0 \), it is a local maximum.

Key Concepts

Critical Points in Calculus

Critical points are where the slope of a function is zero or undefined. These are significant because they can indicate local maxima or minima, or sometimes points of inflection.
To find critical points, you start by taking the first derivative of the function. By setting this derivative to zero, you isolate the values of the independent variable where the function changes behavior.
For our example function, this process gave us the critical points at \( x = 0 \), \( x = -2 \), and \( x = 2 \).
It's essential to assess each critical point further to determine the nature of these points in the context of the function's graph.

Understanding Concavity

Concavity tells us how a function bends or curves. A function is concave up when it bends upwards, like a cup, and concave down when it bends downwards, like a frown.
To find where a function is concave up or down, calculate the second derivative. Where this second derivative is positive, the graph is concave up. Where it is negative, the graph is concave down.
In our exercise, we determined the second derivative as \( y'' = 3x^2 - 4 \). By checking points in intervals around our found roots \( x = \pm \sqrt{\frac{4}{3}} \), we concluded:

• For \( x < -\sqrt{\frac{4}{3}} \), \( y'' > 0 \), which means concave up.
• For \( -\sqrt{\frac{4}{3}} < x < \sqrt{\frac{4}{3}} \), \( y'' < 0 \), indicating concave down.
• For \( x > \sqrt{\frac{4}{3}} \), \( y'' > 0 \), hence concave up again.

Recognizing these patterns can help determine the overall shape of the graph.

Local Maxima and Minima

Local maxima and minima are points on the graph where the function reaches a local highest or lowest value, respectively.
After finding critical points, using the second derivative or by simply substituting values into the function helps further identify these local maxima and minima.
In our solution, at \( x = -2 \) and \( x = 2 \), we verified \( y'' > 0 \), indicating these points are local minima.
Conversely, at \( x = 0 \), because \( y'' < 0 \), the function achieves a local maximum.
Identifying these points is key in sketching graphs and understanding the behavior of functions.