Problem

Find the critical numbers of the function. $ h…

11:16

Problem 35 Medium Difficulty

Find the critical numbers of the function.

$ g(y) = \frac{ y - 1}{ y^2 - y + 1} $

Answer

$$\begin{array}{l}
g(y)=\frac{y-1}{y^{2}-y+1} \Rightarrow \\
g^{\prime}(y)=\frac{\left(y^{2}-y+1\right)(1)-(y-1)(2 y-1)}{\left(y^{2}-y+1\right)^{2}}=\frac{y^{2}-y+1-\left(2 y^{2}-3 y+1\right)}{\left(y^{2}-y+1\right)^{2}}=\frac{-y^{2}+2 y}{\left(y^{2}-y+1\right)^{2}}=\frac{y(2-y)}{\left(y^{2}-y+1\right)^{2}}
\end{array}$$

More Answers

01:45

Wen Z.

Topics

Derivatives

Differentiation

Volume

Book Cover for Calculus: Early Tra…

Calculus: Early Transcendentals

Chapter 4

Applications of Differentiation

Section 1

Maximum and Minimum Values

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Video Transcript

Okay, let's take the derivative using the question rule, which is GF Prime minus half g prime over g squared. In other words, over the denominator squared. We do that. We end up with negative y squared. Plus two. Why? For y squared minus y plus one squared. Simplify this. We end up with negative y squared plus two. Why, doctor? Thus negative y terms. Why mine is too. We end up with two solutions. Weikel zero y equals two.

Amrita B.

Topics

Derivatives

Differentiation

Volume

Book Cover for Calculus: Early Tra…

Calculus: Early Transcendentals

Chapter 4

Applications of Differentiation

Section 1

Maximum and Minimum Values

Top Calculus 2 / BC Educators

Grace H.

Numerade Educator

Catherine R.

Missouri State University

Samuel H.

University of Nottingham

Michael J.

Idaho State University