# functionFind the critical numbers of the function g(y)=(y-5)/(y^2-3y+15)

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### 1 Answer

We'll calculate the first derivative of the function, in order to find the critical values.

The roots of the first derivative are the critical values of the function.

g(y) = (y - 5)/(y^2 - 3y +15)

Since the function is a fraction, we'll calculate the first derivative using quotient rule.

let's recall the quotient rule:

(u/v)' = (u'*v - u*v')/v^2

We'll put u = y - 5 and we'll calculate the first derivative with respect to y.

u' = (y - 5)'

u' = 1

We'll put v = y^2 - 3y +15 and we'll calculate the first derivative with respect to y.

v' = (y^2 - 3y +15)'

v' = 2y - 3

g'(y) = [(y - 5)/(y^2 - 3y +15)]'

g'(y) = [1*(y^2 - 3y +15) - (y - 5)*(2y - 3)]/(y^2 - 3y +15)^2

We'll remove the brackets:

g'(y) = (y^2 - 3y +15 - 2y^2 + 15y - 15)/(y^2 - 3y +15)^2

We'll combine and eliminate like terms:

g'(y) = (-y^2 + 12y)/(y^2 - 3y +15)^2

We'll factorize by y the numerator:

g'(y) = y(12 - y)/(y^2 - 3y +15)^2

Now, we'll calculate the roots of g'(y) = y(12 - y)/(y^2 - 3y +15)^2.

g'(y) = 0

y(12 - y)/(y^2 - 3y +15)^2 =0

Since the denominator is positive for any value of x, only the numerator is cancelling.

y(12 - y)=0

y = 0

12 - y = 0

y = 12

The critical values for g(y) are {0 ; 12}.

The local extreme points are:

g(0) = (0 - 5)/(0^2 - 3*0 +15)

g(0) = -5/15

g(0) = -1/3

g(12) = (12-5)/(12^2 - 3*12 +15)

g(12) = 7/(144 - 36 +15)

g(12) = 7/123

The local extreme points of g(y) are calculated with the critical values and they are: (0,-1/3) and (12, 7/123).