## haleyelizabeth2017 one year ago For which interval(s) is f(x)=x^3+x^2-5x-6 increasing or decreasing?

1. anonymous

increasing maybe

2. haleyelizabeth2017

It's not just an "increasing/decreasing" answer lol I have to find the intervals

3. Owlcoffee

There is something interesting we can actually prove, let's generalize for some point $$a(x_a,y_a)$$ on the plane, we have $$f'(a)>0$$. This must mean that by definition: $$\lim_{x \rightarrow a} \frac{ f(x)-f(x_a) }{ x-x_a }$$. Then by the theorem of conservation of sign we know that there must exist a $$\delta >0$$ that belongs to the interval $$\left| x-a \right|<\delta$$ such that $\frac{ f(x)-f(x_a) }{ x-x_a }>0$ From here we can conclude two things: (1) if $$x<x_a$$ then $$x-x_a<0$$ which implies that $$f(x)-f(x_a)<0$$ then $$f(x)<f(x_a)$$ (2) if $$x>x_a$$ then $$x-x_a>0$$ which implies that $$f(x)-f(x_a)>0$$ then $$f(x)>f(x_a)$$ by the very definition of increasing or decreasing function we can conclude that function "f" is increasing on the point $$a(x_a,y_a)$$. $$if: f'(a)>0 \rightarrow$$ f increases on "a". This process is completely analogical to f'(a)<0. Returning to the excercise: $f(x)=x^3+x^2-5x-6$ What we will do is use the theorem i proved to you earlier, and we will need the derivative of the function in order to apply it: $f'(x)=3x^2+2x-5$ now, we want to know what makes the derivative zero: $3x^2+2x-5=0$

4. haleyelizabeth2017

-5/3 or -1?

5. freckles

@Owlcoffee I don't think @haleyelizabeth2017 knows any calculus

6. haleyelizabeth2017

Lol nope...This would be precalc lol

7. freckles

Calculus is awesome because you can do these questions without a calculator. And it isn't that bad so I guess next year or semester you will get there.

8. Owlcoffee

Great, now you have to study the sign of that interval. That happens to the function if x takes values greater -1? And what happens if x takes values lesser than -5/3? The study of sign looks like this: |dw:1443820022212:dw| This means that for the interval $$(x; + \infty)$$ the function is increasing. for the interval $$(-\frac{ 5 }{ 3 };-1)$$ the function is decreasing. for the interval $$(- \infty ; -\frac{ 5 }{ 3 })$$ the function is increasing. you can verify with your calculator.

9. Owlcoffee

if she's in precalc she must know about derivation, and that's enough to find the variation of a function on the plane @freckles

10. freckles

I helped her with the previous one. She didn't know derivatives if that is what is meant by the word derivation.

11. freckles

So that is why I think these are calculator questions instead.

12. Owlcoffee

I'm sorry, I am from the old school and I did not have those graphing calculators.

13. freckles

I do like old school stuff. I think calculus is fantastic for these questions and they should probably be saved for calculus class.

14. Owlcoffee

I personally believe that teachers shouldn't teach with graphing calculators, and start teaching using their brains.

15. freckles

I totally agree....

16. Owlcoffee

17. freckles

I wasn't trying to put down your way... I was just saying the op is unfamiliar with that way.

18. freckles

Your way actually is the exact way I started on her previous question.

19. Owlcoffee

Don't worry about it, I thought she had the knowledge of derivative. So that's why I started like that.

20. haleyelizabeth2017