Problem 110 - Calculus Volume 1

Question

For the following exercises, find the critical points in the domains of the following functions. $$y=\frac{1}{x-1}$$

Step-by-Step Solution

Answer

There are no critical points in the domain of \( y = \frac{1}{x-1} \).

Step 1: Identify the Function and Domain

We have the function \( y = \frac{1}{x-1} \). This function is undefined at \( x = 1 \) because the denominator becomes zero. Hence, the domain of the function is all real numbers except \( x = 1 \).

Step 2: Find the First Derivative

To find the critical points, we first need to differentiate the function \( y \) with respect to \( x \). Using the quotient rule, \( \frac{d}{dx}\left(\frac{1}{x-1}\right) = -\frac{1}{(x-1)^2} \).

Step 3: Set the Derivative to Zero

Critical points occur where the derivative is zero or undefined. Set \(-\frac{1}{(x-1)^2} = 0\). This situation is impossible because \(-\frac{1}{(x-1)^2}\) can never be zero for any real \( x \). The expression is always negative or undefined.

Step 4: Consider where the Derivative is Undefined

The derivative \(-\frac{1}{(x-1)^2} \) is undefined at \( x = 1 \), where the original function itself is also undefined. Thus, there are no critical points in the domain of the function.

Key Concepts

Domain of a Function

Understanding the domain of a function is essential in solving many mathematical problems. The domain refers to all the possible input values (x-values) for which a function is defined. For the function \( y = \frac{1}{x-1} \), we must exclude the value that would make the denominator zero, resulting in division by zero, which is undefined. - In this case, the critical value is \( x = 1 \). Here, the function becomes \( y = \frac{1}{0} \), which is not allowed mathematically.- Thus, the domain includes all real numbers except \( x = 1 \).Identifying the domain helps in understanding where the function operates continuously and where it might have gaps or undefined points.

First Derivative

The first derivative of a function is a tool used to determine the slope of the tangent line at any point on the curve. It helps in finding critical points, which are points where the function's behavior changes, usually either reaching a maximum or minimum value. For the function \( y = \frac{1}{x-1} \), we use the quotient rule for differentiation to find the first derivative. The result is:- \( \frac{d}{dx}\left(\frac{1}{x-1}\right) = -\frac{1}{(x-1)^2} \)This formulation shows how the rate of change of the function relates to \( x \). It's a vital part of locating critical points, which, generally speaking, occur where this derivative is zero or does not exist. However, in this specific case, since the derivative is negative for all \( x eq 1 \), critical points determined by setting the derivative to zero do not exist.

Undefined Function Values

Undefined function values occur when certain inputs do not result in a valid output, often due to division by zero, discussion of infinity, or square roots of negative numbers. For a mathematical function, it's crucial to identify such points as they have significant impact on the function's behavior. For \( y = \frac{1}{x-1} \), the term \( x-1 \) in the denominator causes the function to be undefined at \( x = 1 \). This is because dividing by zero (as in \( x-1=0 \)) does not yield a real number. - Similarly, the derivative \( -\frac{1}{(x-1)^2} \) is also undefined at \( x = 1 \). This needs to be noted when analyzing critical points, as it implies both the function and its derivative challenge normal operations at this particular value, indicating potential asymptotic behavior or vertical gaps.