Question

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

Answer 1

increasing maybe

Answer 2

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

Answer 3

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:
\[3x^2+2x-5=0\]

Answer 4

-5/3 or -1?

Answer 5

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

Answer 6

Lol nope...This would be precalc lol

Answer 7

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.

Answer 8

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:

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.

Answer 9

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

Answer 10

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

Answer 11

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

Answer 12

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

Answer 13

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

Answer 14

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

Answer 15

I totally agree....

Answer 16

but yeah, I already stated the answer... So... Let's move on!

Answer 17

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

Answer 18

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

Answer 19

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

Answer 20

Sadly not :(

Answer 21

How can someone define increasing or decreasing interval without derivative?...
What's wrong with theachers these days?