Problem 3 - Calculus for Biology and Medicine

Question

Follow Steps 1-10 to find the roots, increasing and decreasing intervals and concave up and concave down intervals of each function along with its behavior at any endpoints (or as \(x \pm \infty)\). Sketch the function, noting on your graph where any local extrema or inflection points are. Determine whether the functions have global maxima and minima, and, if so, note their location on the graph. $$ y=x^{3}-27,-5 \leq x \leq 5 $$

Step-by-Step Solution

Answer

The root is at \(x = 3\), increasing on \([-5, 5]\), concave down on \([-5, 0)\), concave up on \((0, 5]\), global max at \(x=5\), and global min at \(x=-5\).

Step 1: Find the Roots of the Function

The given function is \( y = x^3 - 27 \). To find the roots, set the function equal to zero: \( x^3 - 27 = 0 \). Solving this gives \( x^3 = 27 \), leading to \( x = 3 \). Thus, the function has one real root at \( x = 3 \).

Step 2: Determine Increasing and Decreasing Intervals

Find the first derivative of the function: \( y' = 3x^2 \). Since \( y' = 3x^2 \) is always non-negative and zero only at \( x = 0 \), the function is always non-decreasing on its entire domain \([-5, 5]\). It is specifically increasing everywhere except at \( x = 0 \).

Step 3: Identify Concave Up and Concave Down Intervals

Find the second derivative of the function: \( y'' = 6x \). Determine where \( y'' = 0 \) to find possible inflection points. Solving \( 6x = 0 \) gives \( x = 0 \). For \( x < 0 \), \( y'' < 0 \), indicating concave down. For \( x > 0 \), \( y'' > 0 \), indicating concave up. Hence, the function is concave down on \([-5, 0)\) and concave up on \((0, 5]\).

Step 4: Analyze Behavior at Endpoints and Infinity

Evaluate the function at the endpoints of the given interval: \( y(-5) = (-5)^3 - 27 = -125 - 27 = -152 \) and \( y(5) = (5)^3 - 27 = 125 - 27 = 98 \). As \( x \to -\infty \), \( y \to -\infty \) and as \( x \to +\infty \), \( y \to +\infty \).

Step 5: Examine Local Extrema and Inflection Points

From Step 2, as the derivative does not change sign, there are no local maxima or minima within the interval. From Step 3, the inflection point occurs at \( x = 0 \).

Step 6: Determine Global Maxima and Minima

As analyzed in Steps 4 and 5, the endpoints are the highest and lowest values within the domain. Hence, the global maximum on \([-5, 5]\) is at \( x = 5 \) with \( y = 98 \) and the global minimum is at \( x = -5 \) with \( y = -152 \).

Step 7: Sketch the Graph

Plot the points \((-5, -152)\), \( (3, 0) \), and \( (5, 98) \). Show that from \(x = -5 \) to \(x = 0 \), the curve is concave down, transitioning at the inflection point \(x = 0\) to become concave up from \(x = 0\) to \(x = 5\). Include the global extremum and inflection point on the sketch.

Key Concepts

Roots of a function

When we talk about the roots of a function, we are referring to the values of the variable that make the function equal to zero. In simpler terms, these are the points where the graph of the function crosses the x-axis. For the provided function, \( y = x^3 - 27 \), we find the roots by solving \( x^3 - 27 = 0 \). This simplifies to \( x^3 = 27 \), leading us to \( x = 3 \).
This means the function has one real root at \( x = 3 \), where the graph crosses the x-axis. Finding these points is important because they give us fundamental insights into the function's behavior and help in sketching the graph accurately.

Differentiation

Differentiation allows us to understand how a function changes. Specifically, it helps determine where the function is increasing or decreasing, as well as identifying the rate of change. By finding the first derivative, \( y' \), you can see the slope of the tangent line at any given point. For the function \( y = x^3 - 27 \), the first derivative is \( y' = 3x^2 \).
Since the first derivative \( y' = 3x^2 \) is always non-negative and only equals zero when \( x = 0 \), the function is always non-decreasing over the interval \([-5, 5]\), meaning it's increasing at all points except \( x = 0 \). This implies the function does not have local maximum or minimum within the given domain.

Graph sketching

When sketching the graph of a function, it's important to plot key points and understand the overall shape of the curve concerning different intervals. For instance, with the function \( y = x^3 - 27 \):

• Start by plotting the root at \( (3, 0) \).
• Evaluate the function at the endpoints: \((-5, -152)\) and \((5, 98)\).
• Use the derivatives to determine where the graph is increasing or decreasing and the concavity of the function.

Sketching is not just about plotting points; it’s about understanding how the function behaves between those points.
For this function, sketching involves noting the change in concavity at the inflection point \( x = 0 \), indicating a transition from a concave down to a concave up curve. The graph will appear as a curve that dips down and rises up as it moves from \( x = -5 \) to \( x = 5 \).

Increasing and decreasing intervals

Knowing where a function increases or decreases is crucial for understanding its behavior. This can be determined using the function's first derivative. When the first derivative is positive, the function is increasing; when negative, it is decreasing. For \( y = x^3 - 27 \), as previously noted, \( y' = 3x^2 \) is always non-negative.
Because the derivative never becomes negative within the interval \([-5, 5]\), the function is non-decreasing throughout. It increases on the whole interval except exactly at \( x = 0 \), where the derivative is zero, indicating a temporary halt in increase but no actual decrease. This continuous increase hints at the absence of local maxima or minima.

Concavity and inflection points

Concavity describes how the curve of a function bends. It tells us if the function is curving upwards (concave up) or downwards (concave down). To find this, we look at the second derivative. For \( y = x^3 - 27 \), the second derivative is \( y'' = 6x \).
By solving \( 6x = 0 \), we find that \( x = 0 \) is a potential inflection point. Check intervals around this point: for \( x < 0 \), \( y'' < 0 \) (concave down); for \( x > 0 \), \( y'' > 0 \) (concave up). Thus, \( x = 0 \) marks a shift in concavity from downward to upward. Marking such inflection points reveals where the curvature changes direction, providing insights into the shape and behavior of the graph.