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

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tgl223 | Student, College Freshman | (Level 3) eNoter

Posted November 12, 2010 at 1:12 PM via web

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Find the critical numbers of the function.

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

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted November 12, 2010 at 4:53 PM (Answer #4)

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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).

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neela | High School Teacher | (Level 3) Valedictorian

Posted November 12, 2010 at 5:58 PM (Answer #5)

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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.

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

Posted November 17, 2010 at 5:30 PM (Answer #6)

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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)

 

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