Problem 28 - Calculus for Biology and Medicine

Question

Determine whether the functions have absolute maxima and minima, and, if so, find their coordinates. Find inflection points. Find the intervals on which the function is increasing, on which it is decreasing, on which it is concave up, and on which it is concave down. Sketch the graph of each function. $$ y=x^{4}-2 x^{2}, x \in \mathbf{R} $$

Step-by-Step Solution

Answer

The function has absolute minima at \((-1, -1)\) and \((1, -1)\). Inflection points occur at \( x = \pm \frac{1}{\sqrt{3}} \).

Step 1: Determine Critical Points

To find the critical points, we first compute the derivative of the function \( y = x^{4} - 2x^{2} \). The first derivative is \( y' = 4x^{3} - 4x \). Set this derivative equal to zero to find critical points: \( 4x(x^{2} - 1) = 0 \). Thus, the critical points are \( x = 0, x = 1, x = -1 \).

Step 2: Test Intervals for Increasing/Decreasing

With the critical points \( x = -1, 0, 1 \), test the sign of \( y' \) in the intervals \((-\infty, -1), (-1, 0), (0, 1), (1, \infty)\). For \( x < -1 \), \( y' > 0 \); for \( -1 < x < 0 \), \( y' < 0 \); for \( 0 < x < 1 \), \( y' < 0 \); and for \( x > 1 \), \( y' > 0 \). Hence, the function is increasing on \((-\infty, -1)\) and \((1, \infty)\); decreasing on \((-1, 0)\) and \((0, 1)\).

Step 3: Determine Second Derivative for Concavity

To determine concavity, compute the second derivative: \( y'' = 12x^{2} - 4 \). Set \( y'' = 0 \) to find potential inflection points: \( 12x^{2} = 4 \) gives \( x^{2} = \frac{1}{3} \) or \( x = \pm \frac{1}{\sqrt{3}} \).

Step 4: Test Intervals for Concavity

With potential inflection points \( x = \pm \frac{1}{\sqrt{3}} \), test the sign of \( y'' \) in intervals \((-\infty, -\frac{1}{\sqrt{3}})\), \((-\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}})\), and \((\frac{1}{\sqrt{3}}, \infty)\). For \( x < -\frac{1}{\sqrt{3}} \), \( y'' > 0 \); for \(-\frac{1}{\sqrt{3}} < x < \frac{1}{\sqrt{3}}\), \( y'' < 0 \); and for \( x > \frac{1}{\sqrt{3}} \), \( y'' > 0 \). Thus, the function is concave up on \((-\infty, -\frac{1}{\sqrt{3}}) \cup (\frac{1}{\sqrt{3}}, \infty)\), and concave down on \((-\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}})\).

Step 5: Identify Inflection Points and Extrema

Since the concavity changes at \( x = \pm \frac{1}{\sqrt{3}} \), these are inflection points. Evaluate \( y \) at critical points: \( y(0) = 0 \), \( y(1) = -1 \), and \( y(-1) = -1 \). The absolute minima occur at \( (1, -1) \) and \( (-1, -1) \), and there is no absolute maximum as the function increases to infinity.

Step 6: Sketch the Graph

To sketch the graph, plot the critical points and inflection points. Note that the function is symmetrical along the y-axis due to even-powered terms. The graph falls to minima at points \((-1, -1)\) and \((1, -1)\), increases to infinity both as \(x\to-\infty\) and \(x\to\infty\), and inflects at \(\pm \frac{1}{\sqrt{3}}\).

Key Concepts

Critical Points

Critical points are special numbers in the domain of a function where the function's derivative is either zero or undefined. Think of these as places where the function might change from increasing to decreasing, or vice versa. They help identify peaks, valleys, or flat spots on the graph of the function.

To find critical points for the function given, \( y = x^{4} - 2x^{2} \), we first calculate its first derivative to analyze the rate of change: \( y' = 4x^{3} - 4x \). Setting the derivative to zero helps identify potential turning points: \( 4x(x^{2} - 1) = 0 \). From this equation, we find critical points at \( x = 0 \), \( x = 1 \), and \( x = -1 \). These points serve as key markers when analyzing the behavior of the function.
Manage these points carefully to understand the function's overall shape.

Concavity

Concavity tells us about the shape of the graph—whether it bends upwards or downwards. This concept is crucial to identify whether a point is a maximum, minimum, or neither.

To determine concavity, we use the second derivative of the function. For our function, the second derivative is \( y'' = 12x^{2} - 4 \). We use \( y'' \) to assess whether the graph is bending upwards (concave up) or downwards (concave down):

• When \( y'' > 0 \), the graph is concave up, like a smile.
• When \( y'' < 0 \), the graph is concave down, like a frown.

Finding where \( y'' = 0 \), we solve \( 12x^{2} - 4 = 0 \) to find potential inflection points, leading us to \( x = \pm \frac{1}{\sqrt{3}} \). By considering intervals defined by these points, we determine the regions of concavity, highlighting where the graph changes direction.

Inflection Points

Inflection points are special points on the graph where the concavity changes. This means the graph switches from curving up to curving down, or vice versa.

To find the inflection points, observe where the second derivative changes sign. With the second derivative \( y'' = 12x^{2} - 4 \), we know the sign changes occur at \( x = \pm \frac{1}{\sqrt{3}} \).

• When moving from \(-\infty\) to \(\infty\), the graph changes from concave up to concave down, and back up again at these points.

The values of \( x \) where this happens are the inflection points. In our problem, \( x = -\frac{1}{\sqrt{3}} \) and \( x = \frac{1}{\sqrt{3}} \) are where the graphs transition, making these significant points to note when sketching the function.

Increasing and Decreasing Intervals

Increasing and decreasing intervals tell us where the function is climbing upward or falling downward. Determining these intervals helps us understand the overall direction or trend of a function.

To find these intervals for our function \( y = x^{4} - 2x^{2} \), we look to the first derivative \( y' = 4x^{3} - 4x \). The function increases where \( y' > 0 \) and decreases where \( y' < 0 \). Consider the critical points \( x = -1, 0, 1 \) to divide the real number line into sections:

• Interval \((-\infty, -1)\): \( y' > 0 \), so the function is increasing.
• Interval \((-1, 0)\): \( y' < 0 \), so the function is decreasing.
• Interval \((0, 1)\): \( y' < 0 \), so the function is decreasing.
• Interval \((1, \infty)\): \( y' > 0 \), so the function is increasing again.

These intervals indicate where the function rises or falls, crucial for plotting and understanding how the graph behaves over different sections.