Problem 2 - Calculus for Biology and Medicine

Question

Determine where each function is increasing, decreasing, concave up, and concave down. With the help of a graphing calculator, sketch the graph of each function and label the intervals where it is increasing, decreasing, concave up, and concave down. Make sure that your graphs and your calculations agree. $$ y=x^{2}+5 x, x \in \mathbf{R} $$

Step-by-Step Solution

Answer

The function decreases on \((-\infty, -\frac{5}{2})\), increases on \((-\frac{5}{2}, \infty)\), and is concave up for all \(x \in \mathbb{R}\).

Step 1: Find the First Derivative

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

Step 2: Determine Critical Points

Set the first derivative equal to zero to find the critical points.\[ 2x + 5 = 0 \].Solving for \( x \), we get:\[ x = -\frac{5}{2} \].

Step 3: Test Intervals for Increasing/Decreasing

Choose test points in the intervals \((-\infty, -\frac{5}{2})\) and \((-\frac{5}{2}, \infty)\) to determine if the function is increasing or decreasing:- For the interval \((-\infty, -\frac{5}{2})\), choose \(x = -3\): \[ y'(-3) = 2(-3) + 5 = -1 \]. Since \(y'(-3) < 0\), the function is decreasing in this interval.- For the interval \((-\frac{5}{2}, \infty)\), choose \(x = 0\): \[ y'(0) = 2(0) + 5 = 5 \]. Since \(y'(0) > 0\), the function is increasing in this interval.

Step 4: Find the Second Derivative

To find where the function is concave up or down, compute the second derivative:\[ y'' = \frac{d}{dx}(2x + 5) = 2 \].The second derivative is constant, so it does not change sign but gives us information about concavity.

Step 5: Determine Concavity

Since \( y'' = 2 > 0 \) for all \( x \in \mathbb{R} \), the function is concave up at all points. The function does not have any intervals where it is concave down.

Step 6: Graph and Analyze

Sketch the graph of the function and label the intervals based on analysis:1. From \(( -\infty, -\frac{5}{2}) \), the function is decreasing.2. At \( x = -\frac{5}{2} \), there is a critical point.3. From \(( -\frac{5}{2}, \infty) \), the function is increasing.4. The function is concave up for \( x \in \mathbb{R} \).Use these details to accurately depict the function's behavior on a graph.

Key Concepts

First Derivative

In calculus, the first derivative of a function helps us understand how the function's value changes as we change its input, essentially giving us the slope of the function at any given point. When given a function like \( y = x^2 + 5x \), we can find its first derivative by differentiating it with respect to \( x \). This derivative, \( y' = 2x + 5 \), tells us the slope of the original function at any point \( x \).

### Importance of the First Derivative- It helps determine **increasing or decreasing intervals**: - If \( y' > 0 \), the function is increasing. - If \( y' < 0 \), the function is decreasing.- It's also essential for identifying **critical points**, which are points where the derivative is zero or undefined. Critical points are crucial in understanding the key features of the function's graph.

Critical Points

Critical points occur where the first derivative is equal to zero (0) or where the derivative does not exist. In the function \( y = x^2 + 5x \), we find critical points by setting the first derivative \( y' = 2x + 5 \) to zero. Solving \( 2x + 5 = 0 \), we get \( x = -\frac{5}{2} \). This is the location of our critical point.

### Why Critical Points Matter- They are potential locations for **local maxima or minima**.- By analyzing these points, we can better understand the shape of the graph and make predictions about the function.- Critical points act as boundaries that separate increasing and decreasing intervals of a function. By testing intervals around these points, we can confirm the behavior of the function on different segments.

Concavity

Understanding concavity helps us grasp how a function bends or curves, providing insight into its overall shape and inflection points. For the function \( y = x^2 + 5x \), we find concavity through its second derivative. The second derivative \( y'' = 2 \) is constant, indicating that it doesn't change with \( x \).

### Interpreting Concavity- **Concave up:** If the second derivative \( y'' > 0 \), the function is concave up. It looks like a U shape. - In our case, since \( y'' = 2 \) at all points, the function is concave up for all real numbers \( x \).- **Concave down:** Conversely, if \( y'' < 0 \), the function would be concave down. In our example, this does not occur as the second derivative is positive everywhere.Concavity provides us with crucial data, especially when sketching the curve, as it enhances our understanding of how the function's slope evolves.

Graphing Calculator

A graphing calculator can be a powerful tool in visualizing how mathematical concepts like the first derivative, critical points, and concavity play out on a real graph. When working with functions such as \( y = x^2 + 5x \), using a graphing calculator helps in drawing accurate graphs efficiently.

### Using a Graphing Calculator Effectively- **Plotting the function:** Enter the function equation into the calculator to see its graph. This provides immediate visual confirmation of the function’s behavior.- **Analyzing intervals:** Use the calculator's features to highlight intervals where the function is increasing or decreasing. Confirm these with your derivative calculations.- **Checking concavity:** Evaluate how the curve appears to understand the concavity visually. The upward curve corroborates our earlier finding that the function is concave up across its domain.Employing a graphing calculator complements manual calculations, offering a visual means to better understand complex functions and their characteristics.