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Find the critical points in the domains of the following functions. \(y=\frac{1}{x-1}\)

Short Answer

Expert verified
The function \(y = \frac{1}{x-1}\) has no critical points.

Step by step solution

01

Identify the Domain

First, identify the domain of the function \(y = \frac{1}{x-1}\). The function is undefined where the denominator is zero, so \(x - 1 = 0\) gives \(x = 1\). Therefore, the domain is all real numbers except \(x = 1\).
02

Find the Derivative

Next, find the derivative of the function. Using the quotient rule, we have \(y' = -\frac{1}{(x-1)^2}\).
03

Set Derivative to Zero

To find critical points, set the derivative equal to zero: \(-\frac{1}{(x-1)^2} = 0\). However, this equation has no solutions, as a fraction is never zero unless its numerator is zero, and here, the numerator is \(-1\.\)
04

Determine Where Derivative is Undefined

Instead of being zero, the derivative may not exist at certain points. Check where \(y'\) is undefined. Since \((x-1)^2 = 0\) when \(x = 1\), the derivative is undefined at \(x = 1\). However, \(x = 1\) is not in the domain of \(y\), so it is not a critical point.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Derivatives
When we talk about derivatives, we're discussing the rate at which a function changes. It's a core concept in calculus that helps us understand how one quantity varies relative to another. In our exercise, we needed to find the derivative of the function \( y = \frac{1}{x-1} \). The derivative of a function gives us important information about its behavior, such as where it increases or decreases.

To compute the derivative, we used the Quotient Rule, a technique we'll cover in more depth later. The resulting derivative \( y' = -\frac{1}{(x-1)^2} \) shows us how the slope of the tangent to the curve changes. Notice that the numerator is \(-1\), indicating that the slope is always negative except where the function is undefined. Understanding the derivative helps us find critical points and analyze the behavior of functions.

Critical points occur where the derivative is zero or undefined. In our case, solving \(-\frac{1}{(x-1)^2} = 0\) doesn't provide a solution, because a fraction can't be zero if its numerator isn’t zero. Additionally, we'll see later that the derivative is undefined at \( x = 1 \), which is significant for our function's behavior analysis.
Domain of a Function
The domain of a function tells us all the possible input values (\( x \) values) that the function can accept without breaking any mathematical rules. For the function \( y = \frac{1}{x-1} \), the domain excludes any value that makes the denominator zero, as division by zero is undefined in mathematics.

In our case, setting \( x - 1 = 0 \) reveals that \( x = 1 \) is not part of the domain. Therefore, the domain of \( y \) is all real numbers except \( x = 1 \). This information is crucial when examining the function graphically or analytically, as it indicates where the function has discontinuities or behaves asymptotically.

When evaluating critical points, it's essential to consider whether these points fall within the domain. Critical points outside of the domain cannot be considered valid for the function's behavior analysis. Understanding the domain helps prevent mistakes in interpretation and ensures that we are analyzing the correct intervals.
Quotient Rule
The Quotient Rule is a powerful tool in calculus used to find the derivative of a ratio of two functions. This rule is especially useful when dealing with functions like \( y = \frac{u}{v} \), where both \( u \) and \( v \) are functions of \( x \). The rule is expressed as:
  • \( (\frac{u}{v})' = \frac{v u' - u v'}{v^2} \)
The numerator of the derivative is the difference between the derivative of the numerator times the denominator and the derivative of the denominator times the numerator. The denominator is just the square of the denominator of the original quotient.

For our function \( y = \frac{1}{x-1} \), using the Quotient Rule gives us the derivative \( y' = -\frac{1}{(x-1)^2} \). In this expression:
  • \( u = 1 \) and \( u' = 0 \), since the derivative of a constant is zero.
  • \( v = x - 1 \) and \( v' = 1 \), as the derivative of \( x \) is 1.
Substituting these into the Quotient Rule formula, we determined the correct derivative. The Quotient Rule can simplify and accurately compute derivatives for functions that are ratios, which might otherwise be complicated to differentiate.

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Most popular questions from this chapter

For the following problems, consider a lifeguard at a circular pool with diameter 40 \(\mathrm{m}\) . He must reach someone who is drowning on the exact opposite side of the pool, at position \(C .\) The lifeguard swims with a speed \(v\) and runs around the pool at speed \(w=3 v\) Find a function that measures the total amount of time it takes to reach the drowning person as a function of the swim angle, \(\theta .\)

For the following exercises, use a calculator to graph the function and estimate the value of the limit, then use L'Hôpital's rule to find the limit directly. $$ \lim _{x \rightarrow 0^{+}} \frac{\ln x}{\sin x} $$

For the following exercises, set up, but do not evaluate, each optimization problem. A window is composed of a semicircle placed on top of a rectangle. If you have 20 \(\mathrm{ft}\) of window-framing materials for the outer frame, what is the maximum size of the window you can create? Use \(r\) to represent the radius of the semicircle.

For the following exercises, consider a limousine that gets \(m(v)=\frac{(120-2 v)}{5}\) milgal at speed \(v,\) the chauffeur costs \(\$ 15 / \mathrm{h},\) and gas is \(\$ 3.5 / \mathrm{gal} .\) Find the cheapest driving speed

Use a calculator to graph the function over the interval \([a, b]\) and graph the secant line from \(a\) to \(b .\) Use the calculator to estimate all values of \(c\) as guaranteed by the Mean Value Theorem. Then, find the exact value of \(c,\) if possible, or write the final equation and use a calculator to estimate to four digits. $$ [\mathrm{T}] \quad \mathrm{y}=\sqrt{x+1}+\frac{1}{x^{2}} \text { over }[3,8] $$

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