Problem 44 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Sketch the graph of the function, using the curve-sketching quide of this section. $$ f(t)=3 t^{4}+4 t^{3} $$

Step-by-Step Solution

Answer

The graph of the function \(f(t) = 3t^4 + 4t^3\) has critical points at \(t = 0\) and \(t = -1\) and inflection points at \(t = 0\) and \(t = -\frac{2}{3}\). The graph has no asymptotes, decreases from left to right until \(t = -1\), then increases afterwards. It is initially concave down until \(t = -\frac{2}{3}\), then concave up.

Step 1: Determine the first and second derivatives of the function

First, find the first derivative of the function (which will help us identify critical points): \[ f'(t) = \frac{d}{dt}(3t^4 + 4t^3) = 12t^3 + 12t^2 \] Next, find the second derivative (which will help us identify inflection points): \[ f''(t) = \frac{d^2}{dt^2}(12t^3 + 12t^2) = 36t^2 + 24t \]

Step 2: Determine the critical points and inflection points

Now that we have our first and second derivatives, we need to find the critical points and inflection points. We start with the critical points: To find the critical points, we set the first derivative equal to 0 and solve for t: \[ 12t^3 + 12t^2 = 0 \] Factor out a 12t^2: \[ 12t^2(t + 1) = 0 \] This gives us two critical points, \(t = 0\) and \(t = -1\). Next, find the inflection points by setting the second derivative equal to 0 and solving for t: \[ 36t^2 + 24t = 0 \] Factor out a 12t: \[ 12t(3t + 2) = 0 \] This gives us two inflection points, \(t = 0\) and \(t = -\frac{2}{3}\).

Step 3: Sketch the graph

Now, we are ready to sketch the graph using the information we have collected so far. Begin by plotting the critical points and inflection points: - Critical points: \((0,0)\) and \((-1, -3)\) - Inflection points: \((0,0)\) and \((-2/3, -2.96)\) Sketching the graph, consider the following details: - The graph does not have any asymptotes. - From left to right, the graph is decreasing until \(t = -1\), then it increases afterwards. - The graph initially concaves down until \(t = -\frac{2}{3}\), then it concaves up. Taking these details into account, we can sketch the graph of the given function \(f(t) = 3t^4 + 4t^3\). Thus, the graph of the function has been sketched considering the critical points and inflection points, as well as behavior based on the function and its derivatives.

Key Concepts

Critical Points

Critical points of a function are where the function's slope changes, leading to potential maxima or minima. In our case, after finding the first derivative, we set it equal to zero to find critical points. For the function \( f(t) = 3t^4 + 4t^3 \), the first derivative is given by \( f'(t) = 12t^3 + 12t^2 \). To find where this derivative equals zero, we solve the equation \( 12t^3 + 12t^2 = 0 \). By factoring out \( 12t^2 \), we get \( 12t^2(t + 1) = 0 \). This shows us that the critical points occur at \( t = 0 \) and \( t = -1 \).These values indicate points on the graph where the slope is zero, meaning the graph of the function flattens out at these points. Identifying these points is critical—pun intended—in understanding the function's behavior.

Inflection Points

Inflection points are where the graph of a function changes from concave up to concave down or vice versa. To find these points, we use the second derivative. For our function, the second derivative is \( f''(t) = 36t^2 + 24t \).An inflection point occurs where this second derivative changes sign. We find these by setting the second derivative to zero, solving \( 36t^2 + 24t = 0 \). Factor this equation to \( 12t(3t + 2) = 0 \), giving us the solutions \( t = 0 \) and \( t = -\frac{2}{3} \).At these points, the function's curvature changes direction. For sketching graphs, knowing the inflection points gives us crucial insights into the overall shape of the curve.

Derivative Analysis

Derivative analysis involves determining the behavior of a function using its derivatives. By examining the first derivative \( f'(t) = 12t^3 + 12t^2 \), we can understand where the function is increasing or decreasing.- **Critical Points:** These occur at \( t = 0 \) and \( t = -1 \).- Before \( t = -1 \), when you analyze intervals, notice how the function is decreasing (the slope is negative) until \( t = -1 \).The second derivative \( f''(t) = 36t^2 + 24t \) helps us examine concavity:- **Inflection Points:** Occur at \( t = 0 \) and \( t = -\frac{2}{3} \).- The curve is concave down when \( t < -\frac{2}{3} \) and concave up when \( t > -\frac{2}{3} \).Using both derivatives gives us a comprehensive map of the function's behavior across its domain.

Graphical Behavior

Understanding the graphical behavior of a function is crucial for graph sketching. For \( f(t) = 3t^4 + 4t^3 \), knowing the critical and inflection points frames our sketch.- **Critical Points**: At \( t = 0 \) and \( t = -1 \), the graph flattens.- **Inflection Points**: At \( t = 0 \) and \( t = -\frac{2}{3} \), the curvature of the graph changes.From left to right, the graph decreases until \( t = -1 \). Post \( t = -1 \), it starts increasing. Initially, the graph is concave down, changing to concave up after \( t = -\frac{2}{3} \).This information helps when plotting the graph, providing insights into how the curve should look. It emphasizes the role of critical and inflection points in comprehending a function's visual representation.