Problem 34

Question

Given \(\begin{aligned} f(x) &=\ln x, \quad x \geq 1 \\ &=a x+b, \quad x<1 . \end{aligned}\) Find the constants \(a \& b\) such that \(f^{\prime}(1)\) exists.

Step-by-Step Solution

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Answer
The constants \(a\) and \(b\) such that \(f^{\prime}(1)\) exists are \(a = 1\) and \(b = 0\).
1Step 1: Find Continuity
First, we ensure the function is continuous at \(x = 1\). For that, the function values at \(x = 1\) from the two expressions must be equal, i.e. \( \ln(1) = a\cdot1 + b \). From this, we find that \( b = 0 \), since \( \ln(1) = 0 \).
2Step 2: Find Differentiability
Next, we ensure the function is differentiable at \(x = 1\). This requires that the derivative values at \(x = 1\) from the two expressions are equal. The derivative of \(\ln x\) is \(\frac{1}{x}\), and the derivative of \(ax+b\) is \(a\). Setting these equal at \(x = 1\) gives us: \( \frac{1}{1} = a \). Solving this gives us that \( a = 1 \).

Key Concepts

ContinuityDerivativeLogarithmic Functions
Continuity
In mathematics, continuity is a fundamental property that describes the smoothness and predictability of a function's graph. For a function to be continuous at a particular point, the function's value must match the limit of the function as it approaches that point. This ensures there are no jumps or breaks in the function's graph.

To determine continuity at a specific point, say \( x = 1 \), follow these steps:
  • Calculate the function's limit: Approach the point from both directions (left and right) and ensure the limit of the function is the same.
  • Value and limit match: The function’s value at that point should equal the limit as you approach it from either side.
In the given exercise, the function definition changes at \( x = 1 \):
  • For \( x \geq 1 \), the function is \( f(x) = \ln x \).
  • For \( x < 1 \), the function is \( f(x) = ax + b \).
To ensure continuity, we need \( \ln 1 = a \times 1 + b \). Since \( \ln 1 = 0 \), we find \( b = 0 \), ensuring the left and right sides of the function meet at \( x = 1 \) without any discontinuity.
Derivative
A derivative represents the rate of change of a function concerning its variable. It is a measure of how a function changes as its input changes. In simpler terms, the derivative of a function at a point gives the slope of the tangent line to the graph of the function at that point.

For a function to be differentiable at a particular point, it must be both continuous and have finite derivatives coming from both directions. If the derivatives from the left and right match at the point, the function is differentiable there.

In our exercise, we need to find the derivative of both parts of the function at \( x = 1 \):
  • For \( x \geq 1 \): The derivative of \( \ln x \) is \( \frac{1}{x} \).
  • For \( x < 1 \): The derivative of \( ax + b \) is simply \( a \), since the derivative of a constant \( b \) is zero.
To ensure differentiability, set \( \frac{1}{1} = a \). Solving this, we find \( a = 1 \), meaning the rate of change on both sides of \( x = 1 \) is consistent.
Logarithmic Functions
Logarithmic functions are the inverses of exponential functions, heavily used in areas such as science, engineering, and various fields of mathematics. The natural logarithm, denoted as \( \ln x \), is particularly important because it uses the number \( e \), an irrational and transcendental number approximately equal to 2.71828, as its base.

Key properties of logarithmic functions include:
  • Domain: Only positive real numbers, \( x > 0 \).
  • Range: All real numbers.
  • Basic identity: \( \ln(1) = 0 \).
  • Derivative: The derivative of \( \ln x \) is \( \frac{1}{x} \). This property is crucial in calculus as it helps us find the rate of change of logarithmic functions.
Understanding logarithmic functions is fundamental for solving the exercise since it involves finding points of continuity and differentiation for functions involving both logarithmic expressions and linear terms. In the given problem, recognizing \( \ln 1 = 0 \) is fundamental, ensuring the function's continuity, while \( \frac{1}{x} \) for \( x = 1 \) helps ensure differentiability at that point.