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

3 Answers | Add Yours

changchengliang's profile pic

changchengliang | Elementary School Teacher | (Level 2) Adjunct Educator

Posted on

Critical Numbers

1. Function cuts the horizontal axis when function goes to zero

When g(y)=0 ,  y=5

 

2. Function cuts the vertical axis when y=0

g(0)

= [ 0 - 5] / [ 0^2 - 3y + 15 ]

=  -5 / 15

g(0) =  -1/3

 

3. When y approaches negative infinity

When y approaches negative infinity,

the numerator is dictated by |y|, with |y|>>|5|, hence the overall expression will become negative.

The denominator appoaches y^2 since |y^2| >> |-3y+15|

The entire expression approaches -1/y, which is a negative number very close to zero but never zero!

 

4. When y approaches positive infinity

Likewise, when y approaches positive infinity, the entire expression approaches 1/y, which is a positive number very close to zero but never zero.

 

5. The maximum and minimum points, occuring at the turning points, where the instantaneous gradient of the curve, g'(y) = 0

Let     U= x-5   ;

and    V= x^2 - 3x + 15

g'(y)

= ( V.dU/dy - U.dV/dy ) / V^2

= :

= :

= ( 10x - x^2 ) / (x^2 -3x + 15)^2

 

Turning points occur when the gradient g'(y)=0.

Equating g'(y)=0, we have:

10x - x^2 = 0

or  x.(10-x) = 0

==> turning points occur at

(i) x=0   (see Point no. 2 , where g(0)=-1/3)

(ii) x=10

 

The next question is which is maximum and which is minimum.

To determine that, we now look at the 2nd derivative, g''(y).

Let U = 10x - x^2 ;

and V = (x^2 - 3x + 15)^2

g''(y)

= ( V.dU/dy - U.dV/dy ) / V^2

= :

= :

= [(x^2-3x+15)^2 . (10-2x) - 2(10x-x^2).(x^2-3x+15).(2x-3)] / (x^2-3x+15)^4

g''(0) = 0.0044 (correct to 4 dec place)  > 0   => minimum

g''(10) = -0.0014 (correct to 4 dec place) < 0  => maximum

g(0) = -1/3    (see my Point#2 above)

g(10) = ... = 5/85 = 1/17

Hence the minimum occurs at (0, -1/3) ;

and the maximum occurs at (10, 1/17)

 

neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

Posted on

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

g(y) is function of y , continuous  for all y in the domain (-infinity , infinity).

A critical point of g(y) is a point c for which the g'(c) = 0, for which the tangent is horizontal .

A critical point is also a point where the tagent is vertical.

A critical point is a point where the function is discontinuous.

A critical point is a point where the curve is not differentiable.

Therefore we  find g'(y) and set g'(y) = 0 for the solutions for y. The solutions so found are the critical points.

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

g'(y) = {(y-5)'(y^2-3y+15) -. (y-5)(y^2-3y+15)'}/(y^2-3y+15)^2.

g'(y) = {(y^2-3y+15-(y-5)(2y-3)}/(y^2-3y+15)^2.

g'(y) = (y^2-3y+15 -(2y^2-13y+15)}/(y^2-3y+15).

g'(y) =( -y^2+10y)/(y^2-3y+15)^2.

g'(y) = 0 gives (-y^2+10y) = 0.

y(10-y) = 0. Therefore y = 0 and 10-y = 0, or y = 10.

Therefore , y = 0 or y = 10 are critical points.

Let us examine , whether we get any point of discontinuity in g(y). We lexamine if the denominator y^2-3y+15 becomes = 0:

y^2 - 3y +15  has the discriminant 3^3-4*15 = -51 which is negative. Therefore  y^2-3y+15 > 0 . So there is no point where the denominator becomes 0. So there are no vertical asymptote or vertical tangent also for any finite y = c.

So the critical points are x = 0 and x = 10. The corresponding critical values are of g(y) are g(y) = 0-5/(0^2-3*0+15) = -1/3 and g(10) = (10-5)/(5^2-3*10+45) = 5/40= 1/8.

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

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).

We’ve answered 315,468 questions. We can answer yours, too.

Ask a question