Question

f(x)=x^4 -5x^3 +9x^2
Find the intervals on which f is increasing,
the intervals on which f is decreasing

Answer 1

typo in the question
is it
\[f(x)=x^4 -5x^3-9x^2\] or
\[f(x)=x^4 -5x^3+9x^2\]??

Answer 2

The second one

Answer 3

then the derivative is
\[f'(x)=4x^3-15x^2+18x\] or
\[f'(x)=x(4x^2-15x+18)\] if you are lucky the second part factors

Answer 4

The second part does not factor, so that's my problem

Answer 5

oh actually that makes you even luckier

Answer 6

the second part
\[4x^2-5x+18\] has no real zeros, so it is always positive and you can ignore it

Answer 7

@satellite73 so when it does not factor it means that there are no real zeros?

Answer 8

therefore the derivative is negative if \(x\) is, i.e. negative on \((-\infty, 0)\) and positive otherwise
that make your function decreasing on \((-\infty,0)\) and increasing on \((0,\infty)\)

Answer 9

no

Answer 10

if it does not factor over the integers, you can always find the zeros using the quadratic formula, but if you use it in this case you will see the zeros are complex, so no real zeros

Answer 11

So the stationary point would be (0,0) ?

Answer 12

yes

Answer 13

@satellite73 the critical point would also be 0

Answer 14

you can see it pretty clearly here
http://www.wolframalpha.com/input/?i=x^4+-5x^3+%2B9x^2

Answer 15

as far as i know, stationary point is a synonym for critical point

Answer 16

@satellite73 my teacher didn't let us use the graphing calculator, she only let us use tables. Thanks for all the help :)