Question

please help
show that the function f defined on R by f(x)=x^3-6x^2+9x+4, for all x in R is increasing in

Answer 1

all intervals

Answer 2

Easy, if f'(x) > 0 for all x in R , then f(x) is increasing in all intervals.

Answer 3

i know that f'(x)=3x^2-12x+9

Answer 4

and 3(x^2-4x+3)>0\]

Answer 5

but do not know what next to do

Answer 6

@perl

Answer 7

That should be sufficient proof.

Answer 8

@perl , please help

Answer 9

we can factor that

Answer 10

yes giving us (x-3)(x-1)

Answer 11

3( x - 1)(x-3) > 0

when x < 1 and x < 3 then you have two negatives ,

Answer 12

ok. is it because we have two negatives that is when it increases in all intervals?

Answer 13

you can test numbers. imagine this is a number line:
< ***********(1)*********(3)********** >

Answer 14

lets use zero ,
(0 - 1) ( 0 - 3) = (-1) *(-3) = + 3

Answer 15

< +++++++ (1) ------- (3) ++++++++ >

Answer 16

since f ' (x) > 0 when x < 1 , then f(x) is increasing when x < 1

since f ' (x) > 0 when x > 1 , then f(x) is increasing when x > 1

Answer 17

since f ' (x) > 0 when x > 1 , then f(x) is increasing when x > 1 i do not get this line.

Answer 18

ok. thanks @perl