Question

Find the critical numbers of the function.

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

Short answer

The critical number of the function \(g(y)\) are \(y = 0\) and \(y = 2\).

Step-by-step solution

Given data

The given function is \(g(y) = \frac{{y - 1}}{{{y^2} - y + 1}}\).

Concept of Differentiation

Differentiation is a method of finding the derivative of a function. Differentiation is a process, where we find the instantaneous rate of change in function based on one of its variables.

Determine the derivative of the function

Obtain the first derivative of the given function: \({g^\prime }(y) = \frac{d}{{dy}}\left( {\frac{{y - 1}}{{{y^2} - y + 1}}} \right)\)

Apply quotient rule as shown in equation,

\(\begin{aligned}{c}{g^\prime }(y) &= \frac{{\left( {{y^2} - y + 1} \right)(1) - (y - 1)(2y - 1)}}{{{{\left( {{y^2} - y + 1} \right)}^2}}}\\{g^\prime }(y) &= \frac{{{y^2} - y + 1 - \left( {2{y^2} - 3y + 1} \right)}}{{{{\left( {{y^2} - y + 1} \right)}^2}}}\\{g^\prime }(y) &= \frac{{ - {y^2} + 2y}}{{{{\left( {{y^2} - y + 1} \right)}^2}}}\\{g^\prime }(y) &= \frac{{y(2 - y)}}{{{{\left( {{y^2} - y + 1} \right)}^2}}}\end{aligned}\)

Take \({g^\prime }(y) = 0\) and obtain the critical number.

\(\begin{aligned}{c}\frac{{y(2 - y)}}{{{{\left( {{y^2} - y + 1} \right)}^2}}} &= 0\\y(2 - y) &= 0\end{aligned}\)

Thus, the solutions are y=0 and y=2.

The expression \(\left( {{y^2} - y + 1} \right)\) in the denominator of \({g^\prime }(y)\) cannot be zero as it exists for all the real numbers. Hence, it fails to satisfy the critical number condition.

Hence, the only critical numbers are 0 and 2.

Thus, the only critical numbers of the function \(g(y) = \frac{{y - 1}}{{{y^2} - y + 1}}\) are 0 and 2.