Question

Let g(x)=3x^4-8x^3-30x^2+72x-15

Answer 1

Finding g'(x) or what are we doing?

Answer 2

Oh, smh sorry about that. The instructions then say: Where is g(x) increasing and where is g(x) decreasing? Find the x-values of all relative extrema.

Answer 3

The first step I've done is just finding the derivative idk what's after

Answer 4

Derivative tells us what in terms of this problem?

Answer 5

uhhhh o.0. I don't really understande what my math teacher is doing/saying in class because he's pretty old. So you'd basically be teaching me from scratch :/

Answer 6

Bummer. That's alright. The derivative gives us the slope of the original function at x. In other words, to find the slope of our original function at a point x (we'll say 3) all we have to do is plug x(3) into our derivative. What did you get for g'(x)?

Answer 7

for g'(x) I found 12x^3-24x^2-60x+72

Answer 8

Perfect. Did the problem give a range of values to find where g(x) is increasing or decreasing, or did it just say the whole graph?

Answer 9

I can paste a link of the sample exam that I am studying from if that helps. It's nothing private

Answer 10

Yeah, that's fine.:)

Answer 11

http://math.fau.edu/locke/Courses/MethCalc/2013SpringExam2Sample.pdf

Answer 12

Okay, so back to our problem, we need to find critical points. Critical points are found by finding the zeros of our derivative. In other words, when g'(x)=0
We have 12x^3-24x^2-60x+72=0 the next step is to factor a 12 out, giving us 12(x^3-2x^2-5x+6)=0, which simplified gives us x^3-2x^2-5x+6=0. With me so far?

Answer 13

yes I found that factored "12(x^3....)" part so far

Answer 14

alright, so finding the zeros of that can be a little tricky. We can use the rational root test to help with that though. Rational Root Test is given as p/q where p are your factors of your trailing constant and q is your factors of your leading coefficient. Mumbo jumbo, maybe? p are factors of 1 (coefficient of x^3) and q are factors of 6 (trailing constant) so we get \[\frac{ p }{ q }=\pm1,\pm2,\pm3,\pm6\]
Now we start testing them and find that g(1)=0, which means its a root.
Let's divide by (x-1) a known root now: (x^3 - 2x^2 - 5x + 6)/(x - 1) = x^2 - x - 6
Now we can factor x^2-x-6, like normal and find (x+2)(x-3)

So to summarize we have (x-1)(x+2)(x-3)=0 giving us critical points at 1,-2,and 3.

Answer 15

omg lol

Answer 16

So now to find intervals of increasing/decreasing, we need to define those intervals based on our critical points. We end up with (-inf,-2) (-2,1) (1,3) and (3,+inf). To find if we are increasing/decreasing you pick a point in each interval and evaluate g'(x) for each point. so for our first interval (-inf,-2) let's pick 3. so g'(3)=-24. because g'(3) is negative that tells us the slope of g(x) is negative or DECREASING.
Interval 2: (-2,1) x=0, g'(0)=6, so its positive and therefore g(x) is increasing from -2 to 1. Completing these steps for all 4 intervals gives us where the function is increasing and where it is decreasing...

Answer 17

Ah, I'm sorry. I know it can be really overwhelming at first... let me know if i was unclear on any part.. :/ for the factoring section mathportal.org has some really helpful tools that help evaluate things and explain how to do it yourself really, really well.

Answer 18

Yeah, im just a little stuck on the factoring part of the x^3...stuff

Answer 19

Yeah factoring above x^2 can be really difficult, but practice really helps. Maybe try a few practice problems using this tool... http://mathportal.org/calculators/polynomials-solvers/polynomial-roots-calculator.php

Answer 20

I have to run, but I'll try to get back on a little later tonight or tomorrow. Best of luck.

Answer 21

ok, that is fine, thank you so much for your time however. I'll see how much more of this test I can get done

Answer 22

Were you able to get through factoring and on to setting up your intervals?