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

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)