Question

Find the critical numbers of the function.
g(y)=(y-1)/(y2-y+1)

Answer 1

To find critical numbers of a function, you take the derivative, set it equal to zero, and solve for the variable. For the given function with a denominator that's always positive, we focus on finding where the derivative of the numerator equals zero.

Step-by-step explanation:

Finding the Critical Numbers of a Function

To find the critical numbers of the function g(y)=(y-1)/(y2-y+1), we need to take the derivative of the function and set it equal to zero to solve for y. The critical numbers are the values of y at which the derivative is zero or the derivative is undefined. Since the denominator y^2 - y + 1 is always positive (as it has no real roots), we only need to look for where the derivative of the numerator is equal to zero to find the critical points. Let's go through the steps:

1. Take the derivative of g(y) using the quotient rule, which states that the derivative of a function h(y) = u(y)/v(y) is h'(y) = (v(y)u'(y) - u(y)v'(y)) / (v(y))^2.
2. For g(y), u(y) = y - 1 and v(y) = y^2 - y + 1. The derivative u'(y) is 1, and the derivative v'(y) is 2y - 1.
3. Set the derivative equal to zero and solve for y.

However, we will not perform the actual derivative calculation here. Instead, it is important to provide the methodology for finding critical numbers. Remember that critical numbers are where the derivative equals zero or does not exist.

Answer 2

The critical numbers of the function g(y) = (y-1)/(y²-y+1) are found by taking the derivative, setting it equal to zero, and finding the values of y. Only the values where the derivative is zero or undefined are considered. The critical numbers for this function are 0 and 2.

Step-by-step explanation:

Finding Critical Numbers of a Function

To find the critical numbers of the function g(y) = (y-1)/(y²-y+1), we need to determine where the derivative of g(y) is equal to zero or undefined. Critical numbers are values of y where the function's derivative is zero or not existent, indicating possible local maximums, minimums, or points of inflection.

First, calculate the derivative of g(y) using quotient rule: g'(y) = [(y²-y+1)(1) - (y-1)(2y-1)] / (y²-y+1)². Simplify the numerator and set the derivative equal to zero to find potential critical numbers:

(y²-y+1) - (2y²-2y+1) = 0

-y² + 2y = 0

y(2 - y) = 0

This gives us y = 0 and y = 2 as potential critical numbers. However, we also need to check where the derivative is undefined. Since the denominator of g'(y) is a square of a polynomial, it will never be zero, and thus the derivative will not be undefined.

The critical numbers of g(y) are 0 and 2.