Question

find critical numbers for x^3-6x^2+15

Answer 1

Critical Numbers are when the derivative = 0 right?

so what is the derivative of this function?

Answer 2

yes. so i got 3x^2-12x=0

Answer 3

Alright good...so we have 3x² - 12x = 0

we can factor out a 3 of this equation here...

3(x² - 4x)

can we factor x² - 4x ?

Answer 4

i had 3(x^2-4x) so to factor x^2-4x would it be x(x-4)?

Answer 5

That would be correct...so what values for 'x' would make

3x(x-4) = 0 ?

well...either

3x = 0

or

x - 4 = 0

so what 2 values do we have?

Answer 6

would 4 be one ?

Answer 7

yes 4 would be one value...

what about the

3x = 0 ? what 'x' would make this = 0?

Answer 8

just 0 ?

Answer 9

mmhmm :) and those are the only 2 critical points of this graph!

Answer 10

ahh thank you! now when it says find the open intervals this is increasing or decreasing would it be negative infinity to 0 and 5to infinity ?

Answer 11

Well using correct notation...I believe it should be

[-infinity , 0) && (4 , infinity]

[ and ] means the number is included
( and ) means the number is NOT included

Answer 12

yes im sorry, i meant 4 not 5. and then the relative max would be 4? and the relative min would be 0?

Answer 13

Not quite......Judging from the equation your graph will most likely look something like...



bad drawing...but regardless...relative max would be? and relative min would be?