# Find the critical numbers of the function. g(y) = (y - 2) / (y^{2} − 2y + 4).

**Solution:**

A critical point of a function of f(x), is a value x0 in the domain of f where it is not differentiable or its derivative is zero

i.e. (f ′(x0) = 0).

If f has a local maximum or minimum at c, then c is a critical number of the function f.

Given, g(y) = (y - 2) / (y^{2 }− 2y + 4)

Taking derivative,

g'(y) = (y^{2 }− 2y + 4)(1) - (y - 2) (2y - 2) / (y^{2 }− 2y +4 )^{2}

g'(y) = (y^{2} − 2y + 4) - (2y^{2} - 2y - 4y + 4) / (y^{2} - 2y + 4)^{2}

On simplification,

g'(y) = (y^{2 }- 2y + 4 - 2y^{2 }+ 6y - 4) / (y^{2} - 2y + 4)^{2}

g'(y) = (-y^{2} + 4y ) / (y^{2} - 2y + 4)^{2}

Taking out common term,

g'(y) = y(4 - y) / (y^{2 }- 2y + 4)^{2}

To find critical number, g'(y) = 0

y(4 - y) / (y^{2} - 2y + 4)^{2} = 0

y(4 - y) = 0

y = 0

(4 - y) = 0

y = 4

Therefore, the critical numbers are y = 0 and y = 4.

## Find the critical numbers of the function. g(y) = (y - 2) / (y^{2 }− 2y + 4).

**Summary:**

The critical numbers of the function g(y) = (y - 2) / (y^{2 }− 2y + 4) are y = 0 and y = 4.

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