Problem 7 - Applied Calculus: For Business, Economics, and the Social and Life Sciences

Question

In Exercises 1 through 10 , determine intervals of increase and decrease and intervals of concavity for the given function. Then sketch the graph of the function. Be sure to show all key features such as intercepts, asymptotes, high and low points, points of inflection, cusps, and vertical tangents. $$ g(t)=\frac{t^{2}}{t+1} $$

Step-by-Step Solution

Answer

The function increases on \((-\infty, -2)\) and \((0, \infty)\), and decreases on \((-2, 0)\). It is concave up everywhere but has a vertical asymptote at \( t=-1 \). Major points are intercepts and local extrema.

Step 1: Find the first derivative

To determine intervals of increase and decrease, first find the first derivative of the function. Let \( g(t) = \frac{t^2}{t+1} \). Use the quotient rule: \( g'(t) = \frac{(t+1)(2t) - (t^2)(1)}{(t+1)^2} = \frac{2t^2 + 2t - t^2}{(t+1)^2} = \frac{t^2 + 2t}{(t+1)^2}. \)

Step 2: Find critical points

Next, set the first derivative equal to zero to find critical points: \( \frac{t^2 + 2t}{(t+1)^2} = 0 \). This simplifies to \( t(t + 2) = 0 \), giving critical points \( t = 0 \) and \( t = -2 \).

Step 3: Determine the intervals of increase and decrease

Determine the sign of the first derivative on the intervals defined by the critical points: Test intervals: - For \( t < -2 \): Choose \( t = -3 \), \( g'(-3) = \frac{(-3)^2 + 2(-3)}{(-3+1)^2} = \frac{9 - 6}{4} > 0 \) (increasing). - For \( -2 < t < 0 \): Choose \( t = -1 \), \( g'(-1) = \frac{(-1)^2 + 2(-1)}{(-1+1)^2} = \frac{1 - 2}{1} < 0 \) (decreasing). - For \( t > 0 \): Choose \( t = 1 \), \( g'(1) = \frac{(1)^2 + 2(1)}{(1+1)^2} = \frac{1 + 2}{4} > 0 \) (increasing).

Step 4: Find the second derivative

To determine the intervals of concavity, find the second derivative. Use the quotient rule for \( g'(t) = \frac{t^2 + 2t}{(t+1)^2} \): Let \( u = t^2 + 2t \) and \( v = (t+1)^2 \), then \( g''(t) = \frac{v(u') - u(v')}{v^2} \). Calculating, \( g''(t) = \frac{(t+1)^2(2t + 2) - (t^2 + 2t)2(t+1)}{(t+1)^4} \) simplifies to \( g''(t) = \frac{2t^2 + 4t + 2 - 2t^2}{(t+1)^3} = \frac{2t + 2}{(t+1)^3} \).

Step 5: Find points of inflection

Set the second derivative equal to zero \( \frac{2(t + 1)}{(t+1)^3} = 0 \). Calculating, it gives no additional critical points. Thus, check the sign of the second derivative: - For \( t < -1 \): Choose \( t = -2 \), \( g''(-2) = \frac{2(-2 + 1)}{(-2 + 1)^3} = \frac{-2}{-1} > 0 \) (concave up). - For \( t > -1 \): Choose \( t = 0 \), \( g''(0) = \frac{2(0 + 1)}{(0+1)^3} = 2 > 0 \) (concave up). Hence, there is no change in concavity.

Step 6: Sketch the graph and identify key features

Identify the key features for sketching the graph: - Intercepts: The y-intercept is at \( g(0) = 0 \). There are no x-intercepts since the x-values are divided by a positive denominator. - Asymptotes: There is a vertical asymptote at \( t = -1 \). - High/Low points: A local minimum at \( t = -2 \) as \( g(-2) = \frac{4}{-1} = -4 \), and a local maximum at \( t = 0 \). - Points of inflection: None (no intervals where concavity changes). Sketch with these features.

Key Concepts

Intervals of Increase

In calculus, intervals of increase refer to the sections of a graph where the function is getting larger as you move from left to right. To find these intervals, we look at the first derivative. For the function \( g(t) = \frac{t^2}{t+1} \), its first derivative is \( g'(t) = \frac{t^2 + 2t}{(t+1)^2} \). By setting \( g'(t) \) greater than zero, we can determine where the function is increasing.

From the solution, we found that the function increases for:

• \( t < -2 \)
• \( t > 0 \)

This means the function is rising in these intervals. Understanding this helps us sketch accurate graphs and predict the behavior of the function over different intervals.

Intervals of Decrease

Intervals of decrease are where the function is declining as you move from left to right on the graph. Again, we use the first derivative, \( g'(t) = \frac{t^2 + 2t}{(t+1)^2} \). By setting \( g'(t) \) less than zero, we find where the function is decreasing.

The solution tells us the function decreases in the interval:

• \( -2 < t < 0 \)

These intervals show us exactly where the function descends, which is crucial for understanding the overall shape of the graph.

Intervals of Concavity

Concavity refers to whether a graph bends upwards or downwards. We determine concavity using the second derivative. For our function, the second derivative is \( g''(t) = \frac{2t + 2}{(t+1)^3} \).

By examining the sign of \( g''(t) \), we can find intervals of concavity:

• For \( t < -1 \): The function is concave up.
• For \( t > -1 \): The function is also concave up.

Since the concavity does not change, there are no intervals of concave down in this case.

Critical Points

Critical points are where the first derivative equals zero or where the first derivative is undefined. These points help identify where the function might have a maximum, minimum, or a point of inflection.

The critical points for the function \( g(t) = \frac{t^2}{t+1} \) are found by setting \( g'(t) = 0 \). Solving \( \frac{t^2 + 2t}{(t+1)^2} = 0 \) gives:

• \( t = 0 \)
• \( t = -2 \)

These critical points are important as they mark potential local maximum or minimum values on the graph.

First Derivative

The first derivative gives us information about the rate of change of the function. It tells us where the function is increasing or decreasing. For our function, the first derivative, using the quotient rule, is: \( g'(t) = \frac{t^2 + 2t}{(t+1)^2} \).

This first derivative is crucial for finding critical points and determining intervals of increase and decrease. If \( g'(t) > 0 \), the function is increasing. If \( g'(t) < 0 \), the function is decreasing.

Second Derivative

The second derivative provides information about the concavity of the function. It tells us whether the function is curving upwards or downwards. For our function, the second derivative is: \( g''(t) = \frac{2t + 2}{(t+1)^3} \).

By checking the sign of \( g''(t) \), we determine concavity. If \( g''(t) > 0 \), the graph is concave up. If \( g''(t) < 0 \), the graph is concave down.

Asymptotes

Asymptotes are lines that the graph approaches but never actually touches. For rational functions like \( g(t) = \frac{t^2}{t+1} \), vertical asymptotes typically occur where the denominator is zero but the numerator is not zero.

For our function, there is a vertical asymptote at

• \( t = -1 \)

It’s important to note this when sketching the graph because it indicates where the function will tend to infinity.

Points of Inflection

Points of inflection occur where the concavity of the function changes. They are found by setting the second derivative equal to zero. In our case, the second derivative is \( g''(t) = \frac{2t + 2}{(t+1)^3} \).

Setting \( g''(t) = 0 \) gives no new critical points, so:

• There are no points of inflection as the concavity doesn't change.

Understanding the absence or presence of inflection points helps in accurately sketching and interpreting the graph.