Problem 2 - Calculus for Biology and Medicine

Question

Use the first derivative test and the second derivative test to determine where each function is increasing, decreasing, concave up, and concave down. You do not need to use a graphing calculator for these exercises. $$ y=x^{2}+5 x, x \in \mathbf{R} $$

Step-by-Step Solution

Answer

The function is decreasing for \( x < -\frac{5}{2} \), increasing for \( x > -\frac{5}{2} \), and concave up for all \( x \).

Step 1: Find the First Derivative

To determine where the function is increasing or decreasing, first find the first derivative of the function. Given the function: \[ y = x^2 + 5x \]The first derivative is computed as:\[ y' = \frac{d}{dx}(x^2 + 5x) = 2x + 5 \]

Step 2: Set the First Derivative to Zero

Next, find the critical points where the function could change from increasing to decreasing or vice versa. Set the first derivative equal to zero and solve for \( x \):\[ 2x + 5 = 0 \]Subtract 5 from both sides:\[ 2x = -5 \]Divide by 2:\[ x = -\frac{5}{2} \]

Step 3: Test the Intervals around Critical Points

Determine whether the function is increasing or decreasing by testing values in the intervals created by the critical point \( x = -\frac{5}{2} \). Check intervals to the left and right of \( -\frac{5}{2} \):For \( x < -\frac{5}{2} \) (e.g., \( x = -3 \)):\[ y' = 2(-3) + 5 = -6 + 5 = -1 \] (function is decreasing)For \( x > -\frac{5}{2} \) (e.g., \( x = 0 \)):\[ y' = 2(0) + 5 = 5 \] (function is increasing)

Step 4: Find the Second Derivative

Next, find the second derivative to determine concavity. The second derivative is:\[ y'' = \frac{d}{dx}(2x + 5) = 2 \]

Step 5: Analyze Concavity Using Second Derivative

The second derivative is constant and positive (\( y'' = 2 \)), meaning the function is always concave up for all \( x \) in \( \mathbf{R} \). Since \( y'' > 0 \), there are no inflection points.

Key Concepts

Second Derivative Test

The second derivative test helps determine the concavity of a function and identify potential points of inflection. By analyzing the sign of the second derivative, we figure out if a function is concave up or concave down in certain intervals. In the given exercise, for the function \[ y = x^2 + 5x \], the second derivative is constant and equal to 2:

\[ y'' = 2 \].This means that the second derivative is positive on the entire real number line. Here's how to interpret this:

• Since \( y'' > 0 \), the graph of the function is concave up at every point on the real line.
• Because it's always concave up, there can't be any inflection points, where the concavity might change.

In summary, with a second derivative that doesn't change sign, this part of the test provides a straightforward conclusion about the nature of the function's graph.

Increasing and Decreasing Functions

To determine where a function is increasing or decreasing, the first derivative is used. It tells us the slope of the tangent line to the function at any given point. For the function \[ y = x^2 + 5x \], we found the first derivative:

\[ y' = 2x + 5 \].
By setting this derivative equal to zero, we locate the critical points, which indicate where the function could switch from increasing to decreasing. After solving:

\[ 2x + 5 = 0 \],
we find the critical point:\[ x = -\frac{5}{2} \].Testing the intervals around \( x = -\frac{5}{2} \):

• For \( x < -\frac{5}{2} \), such as at \( x = -3 \), the derivative \( y' = -1 \), indicating the function is decreasing.
• For \( x > -\frac{5}{2} \), such as at \( x = 0 \), the derivative \( y' = 5 \), indicating the function is increasing.

Thus, the function decreases on the interval \((-\infty, -\frac{5}{2})\) and increases on \((-\frac{5}{2}, \infty)\). This clear separation of intervals based on the sign of the derivative helps map out the behavior of the function.

Concavity

Concavity deals with the direction in which a function curves, which can be determined by the second derivative. If the second derivative is positive, the function is concave up, resembling the shape of a cup. If negative, it's concave down, like an upside-down cup.

For the exercise function \[ y = x^2 + 5x \], the second derivative is:\[ y'' = 2 \], which is constant and positive. Therefore:

• Run through the values of \( x \) and always find \( y'' = 2 \, (\text{positive}) \), implying the function is always concave up.

This means the graph is shaped like an upward-opening parabola at every point. In easier terms, think of rolling a ball; it would always roll into the curve, reinforcing the idea of concave up. Since the second derivative doesn't flip signs, no changes in concavity occur, making this graph consistent throughout.

Critical Points

Identifying critical points is crucial for understanding where functions might change their behavior from increasing to decreasing, or vice versa. These points occur where the first derivative is zero or undefined. For \[ y = x^2 + 5x \], we calculate:

\[ y' = 2x + 5 \].Set it equal to zero:\[ 2x + 5 = 0 \].
Solve to get:\[ x = -\frac{5}{2} \].This critical point at \( x = -\frac{5}{2} \) acts as a pivot point for the function's behavior. Here's what happens around the critical point:

• To the left (\( x < -\frac{5}{2} \)), the function decreases because the derivative is negative.
• To the right (\( x > -\frac{5}{2} \)), the function increases because the derivative is positive.

While critical points don't necessarily tell us about peaks or troughs directly, they help in predicting shifts in the graph's movement, particularly for polynomial functions.