Question

A function \(f(x)\) is given. (a) Give the domain of \(f\). (b) Find the critical numbers of \(f\). (c) Create a number line to determine the intervals on which \(f\) is increasing and decreasing. (d) Use the First Derivative Test to determine whether each critical point is a relative maximum, minimum, or neither.. \(f(x)=\frac{1}{x^{2}-2 x+2}\)

Short answer

Domain: all real numbers. Critical number: 1. Increasing: \((-\infty, 1)\); decreasing: \((1, \infty)\). Max at x=1.

Step-by-step solution

Finding the Domain

The function is given as \( f(x) = \frac{1}{x^2 - 2x + 2} \). To find the domain, we need to identify values of \( x \) for which the denominator is zero, as the function is undefined at these points. The denominator can be rewritten as \( (x-1)^2 + 1 \). Since \((x-1)^2\) is always non-negative and \(+1\) ensures it is positive, the expression \( x^2 - 2x + 2 \) is never zero. Thus, the domain of \( f \) is all real numbers: \( \mathbb{R} \).

Finding the Critical Numbers

Critical numbers occur where the first derivative is zero or undefined. First, we find \( f'(x) \). Using the quotient rule, \( f'(x) = \frac{d}{dx}\left(\frac{1}{x^2 - 2x + 2}\right) = \frac{0 \cdot (x^2 - 2x + 2) - 1 \cdot (2x - 2)}{(x^2 - 2x + 2)^2} = \frac{-(2x - 2)}{(x^2 - 2x + 2)^2} = \frac{2 - 2x}{(x^2 - 2x + 2)^2} \). Setting the numerator equal to zero, \( 2 - 2x = 0 \), gives \( x = 1 \). Thus, \( x = 1 \) is a critical number.

Creating a Number Line for Intervals of Increase/Decrease

We test the sign of the first derivative \( f'(x) = \frac{2 - 2x}{(x^2 - 2x + 2)^2} \) on intervals determined by the critical number \( x = 1 \). Choose test points from intervals \((-\infty, 1)\) and \((1, \infty)\). For \( x = 0 \) (in \((-\infty, 1)\)), \( f'(0) = \frac{2}{4} > 0 \), so \( f \) increases there. For \( x = 2 \) (in \((1, \infty)\)), \( f'(2) = \frac{-2}{4} < 0 \), so \( f \) decreases there.

Applying the First Derivative Test

To determine the nature of the critical point at \( x = 1 \), we apply the First Derivative Test. Since \( f'(x) \) changes from positive to negative at \( x = 1 \), \( f \) has a relative maximum at \( x = 1 \).

Key concepts

Function Domain

When we talk about the domain of a function, we are focusing on the set of all possible input values (in this case, values for \( x \)) that make the function work without any hiccups. For the function \( f(x) = \frac{1}{x^2 - 2x + 2} \), finding the domain means uncovering all the \( x \) values where the function is defined without causing division by zero—specifically, ensuring the denominator doesn't become zero.

In this particular example, the expression in the denominator \( x^2 - 2x + 2 \) can be rewritten as \( (x-1)^2 + 1 \). Now, think of \( (x-1)^2 \): it's always non-negative (zero or positive), right? That's simply because any number squared will never be negative.

• Adding 1 to a non-negative number always keeps it positive, meaning \( x^2 - 2x + 2 \) is also always positive.
• This nifty little arrangement means that ◆the denominator never reaches zero.

Therefore, there aren't any restrictions on the \( x \) values for our function, making the domain all real numbers, symbolically \( \mathbb{R} \).

First Derivative Test

The First Derivative Test is a powerful tool for determining the behavior of a function at critical points, which are points where the function's derivative is zero or undefined. For the function \( f(x) = \frac{1}{x^2 - 2x + 2} \), we first found the derivative using the Quotient Rule. The resulting first derivative was \( f'(x) = \frac{2 - 2x}{(x^2 - 2x + 2)^2} \).

Critical numbers are then found where this derivative equals zero or is undefined. Setting the numerator, \( 2 - 2x = 0 \), gives us the critical number \( x = 1 \). With critical points identified, the First Derivative Test will help us decide if these points correspond to a local maximum, minimum, or neither.

• By testing the sign of the derivative around \( x = 1 \), for instance, using a point from each interval like \( x = 0 \) and \( x = 2 \), we can see how the sign changes.
• If \( f'(x) \) transitions from positive to negative, like it does at \( x = 1 \), it indicates that the function peaks at this point, resulting in a local maximum.

Applying this test confirms that there is a relative maximum at \( x = 1 \) for the function in question. This analysis is not only an assertion of whether a point is a maximum or minimum but also observers the behavior of a function across different intervals.

Quotient Rule

The Quotient Rule is a method used in calculus to find the derivative of a function that is the ratio of two differentiable functions. For our given function \( f(x) = \frac{1}{x^2 - 2x + 2} \), recognizing its structure as a fraction is key to employing the Quotient Rule.

The rule itself states that if you have a function \( g(x) = \frac{u(x)}{v(x)} \), the derivative is found using:\[ g'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} \]

• In our function, \( u(x) = 1 \) with its derivative \( u'(x) = 0 \), and \( v(x) = x^2 - 2x + 2 \) with its derivative \( v'(x) = 2x - 2 \).
• Substituting in these values gives us the formula for the derivative \( f'(x) = \frac{0 imes (x^2 - 2x + 2) - 1 imes (2x - 2)}{(x^2 - 2x + 2)^2} \) which simplifies to \( \frac{2 - 2x}{(x^2 - 2x + 2)^2} \).

This approach helps unravel the behavior of our function and enables further analysis like finding critical points for understanding where the function increases or decreases.