x^3-5x^2-6x less than or equal to 0
answer is -1,6 need help solving.

Question

x^3-5x^2-6x less than or equal to 0
answer is -1,6 need help solving.

anonymous · Answer

what do you need on

anonymous · Answer

help on

anonymous · Answer

I have the answers but don't know the process of solving it

zzr0ck3r · Answer

$x^3-5x^2-6x\le 0\x(x^2-5x-6)\le 0\x(x+1)(x-6)\le0$

So we have roots at $0,-1$ and $6$.

note that this thing has a positive coefficient on the first term, and it is a polynomial of degree three, so it looks like $x^3$ for its end behavior, in other words it starts out negative. 

So since all of our roots are odd powers, it actually crosses the x-axis at each root (unlike $x^2,x^4,...$), so we start out negative, then cross the x-axis at $x=-1$ and so we are positive now, then it crosses back over at $x=0$ so now we are negative again, then it crosses back over the x-axis again at $x=6$ then it remains positive from then on out (because there are no more roots).

So finally to answer the question.

The function is negative on $(-\infty, -1)\cup(0,6)$

Note that we use open brackets because it is NOT negative at the roots; it's $0$ at the roots.

zzr0ck3r · Answer

woops, I forgot we were doing $\le$ so it actually would be 

$$(-\infty, -1]\cup[0,6]
 
$$

campbell_st · Answer

Just sketch it

campbell_st · Answer

And look at the parts below the x axis.... Then define the inequalities

zzr0ck3r · Answer

I agree! if you can use a calculator, or a graphing device. I normally work under the assumption that we cant use these things. You would need all the info I gave to sketch it without software?

campbell_st · Answer

well if you know the zeros and knowing the coefficient of the leading term is positive then the sketch is basically 

|dw:1417630826083:dw|

now its easy to define the inequality without a lot of complicated work.