94.4k views
3 votes
Critical numbers occur where g'(y) equals 0 or is undefined. g'(y) is undefined where the quadratic y2 − 2y + 4 in the denominator is 0. So, g'(y) is undefined for the following values. (Enter your answers as a comma-separated list. If an answer does not exist, enter DNE.)

1 Answer

4 votes

Answer:

There are no real values of y for which g'(y) is undefined.

Explanation:

We are given the following in the question:


g(y) = y^2 - 2y +4

We have to find values for which g'(y) is undefined.


g'(y) = 2y-2

Putting g(y) = 0, we get:


y^2 - 2y +4 = 0\\y^2 -2y +1+3 = 0\\(y-1)^2=-3\\y-1=√(-3)\\y-1 = i\sqrt3\\y = 1 \pm i\sqrt3

Since those are not real numbers, there are no real numbers where g'(y) is undefined.

Thus, there are no real values of y for which g'(y) is undefined.

User Cybersupernova
by
7.8k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.