What are the critical numbers of g(y)?

Question

#(y-5)/(y^2 -3y + 15)#

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Answer 1 · 2017-11-08T22:06:35

#y=0# and #y=10#

Differentiate using the quotient rule:

#(dg)/dy = ( (y^2-3y+15) d/dy (y-5) - (y-5) d/dy (y^2-3y+15))/(y^2-3y+15)^2#

#(dg)/dy = ( (y^2-3y+15) - (y-5) (2y-3))/(y^2-3y+15)^2#

#(dg)/dy = ( (y^2-3y+15) - (2y^2 -3y -10y+15))/(y^2-3y+15)^2#

#(dg)/dy = ( y^2-3y+15 - 2y^2 +13y -15)/(y^2-3y+5)^2#

#(dg)/dy = -( y^2-10y )/(y^2-3y+5)^2#

The critical points are the values of #y# for which the derivative is zero, that is the roots of the equation:

#( y(y-10) )/(y^2-3y+5)^2 = 0#

For the fraction to be zero the numerator must be zero:

#y(y-10)= 0#

For both solutions #y=0# and #y=10# the denominator is non zero so they are critical points of the functions.