Problem 36 - Applied Calculus: For Business, Economics, and the Social and Life Sciences

Question

\(g(t)=\frac{\ln (t+1)}{t+1}\)

Step-by-Step Solution

Answer

Domain is \((-1, \infty)\), no singularities, and the function is continuous and differentiable for \(t > -1\).

Step 1: Understand the Function

Given the function: \[g(t) = \frac{\ln (t+1)}{t+1}\]This function is a combination of the natural logarithm and a polynomial expression.

Step 2: Find the Domain of the Function

The function is defined for all values of \(t\) such that the argument of the logarithm is positive. Since \(t + 1\) must be greater than zero, \(t > -1\). The domain of the function is thus \((-1, \infty)\).

Step 3: Investigate Potential Singularities

Identify points where the denominator could potentially cause a singularity. In this case, at \(t = -1\) the term \(t + 1 = 0\) leading to division by zero. However, \(t = -1\) is not in the domain, hence no singularity within the domain.

Step 4: Check for Continuous and Differentiable Properties

Check if the function is continuous and differentiable in its domain. Since both \(\ln(t+1)\) and \(t+1\) are continuous and differentiable for \(t > -1\), their quotient is also continuous and differentiable over the same interval. This confirms that the function has no discontinuities in its domain.

Key Concepts

Domain of a Function

The domain of a function refers to all the possible inputs (or values) for which the function is defined. For the function given in the exercise, \(g(t) = \frac{\ln(t+1)}{t+1}\), we need to ensure the input values make the expression inside the logarithm, ln(t+1), valid (positive) and avoid division by zero.

The logarithmic function, \(\ln(t+1)\), is defined only for \(t+1 > 0\). This means that \(t > -1\).

Additionally, the expression \(\frac{\ln(t+1)}{t+1}\) requires t ≠ -1 since division by zero is undefined. Combining both conditions, the domain of the function g(t) is; (-1, ∞). This interval represents all the possible values of t for which the function exists.

Continuous Function

A continuous function is a function that has no breaks, holes, or gaps. In other words, you can draw the graph of the function without lifting your pen off the paper.

For the function \(g(t) = \frac{\ln(t+1)}{t+1}\) to be continuous, both \(\ln(t+1)\) and \(t+1\) must themselves be continuous.

The natural logarithm \(\ln(t+1)\) is continuous for \(t > -1\), and \(t+1\) is a polynomial, which is always continuous.

Since the quotient of two continuous functions is also continuous (as long as the denominator is not zero), \(g(t)\) is continuous for all t in its domain, (-1, ∞). This means there are no interruptions in the graph of g(t) within this interval.

Differentiable Function

A differentiable function is one that has a derivative at every point in its domain. The derivative is a measure of how the function changes at any given point.

For \(g(t)\) to be differentiable, the functions ln(t+1) and t+1 must be differentiable as well.

The natural logarithm function \(\ln(x)\) is differentiable wherever it is defined, and polynomials like \( t+1 \) are differentiable everywhere.

Therefore, since both \(\ln(t+1)\) and \(t+1\) are differentiable, and the quotient rule allows us to differentiate their quotient without issue within the domain of \((-1, \textbackslash infinity)\), we conclude that \(g(t)\) is differentiable for all \(t > -1\)

Natural Logarithm

The natural logarithm, represented as ln(x), is a fundamental mathematical function. It is the inverse of the exponential function \(e^x\). The natural logarithm of a number y is the power to which e (approximately 2.71828) must be raised to get y.

For example, ln(7.389) is approximately 2 because \(e^2\) is about 7.389.

ln(x) has several key properties:

• ln(1) = 0 because \(e^0 = 1\)
• ln(e) = 1 because \(e^1 = e\)
• ln(ab) = ln(a) + ln(b)
• ln(a/b) = ln(a) - ln(b)
• ln(a^b) = b ln(a)

For the function \(g(t) = \frac{\ln(t+1)}{t+1}\), the natural logarithm applies to \((t + 1)\), transforming it in a way that is defined and smooth for \(t > -1\). This introduces an interesting tail effect to the function plot.