Where f(x) increases/decreases

[Deckler]

The Question:
1) f(x) = x[sup:30prntn5]3[/sup:30prntn5]-9x[sup:30prntn5]2[/sup:30prntn5]+24x+4
Find:
a) local extrema
b) Inflection Points
c.1) f(x) increasing
c.2) f(x) decreasing
d.1) concave up
d.2) concave down

My work:

y'= 3x[sup:30prntn5]2[/sup:30prntn5]-18x+24 OR (3x-12)(x-2)
y''=6x-18

a) Ran y' through quadratic equation, got 2,4. Graph confirms.

b)
6x-18=0
6x=18
x=3, Graph confirms.

c.1)
(3x-12)(x-2)>0
3x-12>0
3x>12 x>4, graph confirms.
x-2>0
x>2. Uh, what? That should be x<2.

c.2) Same as above, but reversed, getting x>2 (graph confirms) and x>4 (What?)

d.1)
6x-18>0
6x>18
x>3, graph confirms.

d.2)
6x-18<0
6x<18
x<3, graph confirms.

What am I screwing up on c.1 and c.2?

 

[Deleted member 4993]

The Question:
1) f(x) = x[sup:23kg5d2e]3[/sup:23kg5d2e]-9x[sup:23kg5d2e]2[/sup:23kg5d2e]+24x+4
Find:
a) local extrema
b) Inflection Points
c.1) f(x) increasing
c.2) f(x) decreasing
d.1) concave up
d.2) concave down

My work:

y'= 3x[sup:23kg5d2e]2[/sup:23kg5d2e]-18x+24 OR (3x-12)(x-2)
y''=6x-18

a) Ran y' through quadratic equation, got 2,4. Graph confirms.

b)
6x-18=0
6x=18
x=3, Graph confirms.

c.1)
(3x-12)(x-2)>0

f'(x) changes sign at x =2 and x=4

So you need to look at the sign of f(x) at these three regions

(-inf, < x < 2) f'(x) is positive and f(x) is increasing

(2 < x < 4) f'(x) is negative and f(x) is decreasing

(4 < x < inf.) f'(x) is positive and f(x) is increasing

3x-12>0
3x>12 x>4, graph confirms.
x-2>0
x>2. Uh, what? That should be x<2.

c.2) Same as above, but reversed, getting x>2 (graph confirms) and x>4 (What?)

d.1)
6x-18>0
6x>18
x>3, graph confirms.

d.2)
6x-18<0
6x<18
x<3, graph confirms.

What am I screwing up on c.1 and c.2?

 

[Deckler]

Yeah, I know my answers are x<2, x>4 for c.1 and 2<x<4 for c.2 thanks to my graph, but how do I find that out? My calc teacher is notoriously bad, and I've been pretty much learning from the internet. (Bad as in, him telling us AFTER the test that, oh by the way, f'(x) is the velocity, and f''(x) is the acceleration of f(x). That's how you solved those problems the entire test was based on and you had never seen before in your life. He honestly never told us that, we saw these velocity/acceleration word problems on the test and were like 'We haven't covered this...")

Rant aside, what's the hard math to get that answer?

 

[Deleted member 4993]

Yeah, I know my answers are x<2, x>4 for c.1 and 2<x<4 for c.2 thanks to my graph, but how do I find that out? My calc teacher is notoriously bad, and I've been pretty much learning from the internet. (Bad as in, him telling us AFTER the test that, oh by the way, f'(x) is the velocity, and f''(x) is the acceleration of f(x). That's how you solved those problems the entire test was based on and you had never seen before in your life. He honestly never told us that, we saw these velocity/acceleration word problems on the test and were like 'We haven't covered this...")

Rant aside, what's the hard math to get that answer?

There is no other hard math - except what I told you in the previous post.

 

[Deckler]

Err, let me restate. My entire knowledge of finding increasing and decreasing consists of f'(x) > 0 and f'(x) < 0.

I haven't the slightest clue what you mean by sign changes in regions.

 

[Deleted member 4993]

Err, let me restate. My entire knowledge of finding increasing and decreasing consists of f'(x) > 0 and f'(x) < 0.

I haven't the slightest clue what you mean by sign changes in regions.

Put on your thinking cap!!

and read my response carefully.

What happens when f'(x) changes sign - it becomes positive from negative (or vice versa).

 

[BigGlenntheHeavy]

\(\displaystyle f(x) \ = \ x^{3}-9x^{2}+24x+4.\)

\(\displaystyle f \ ' \ (x) \ = \ 3x^{2}-18x+24 \ = \ 3(x^{2}-6x+8) \ = \ 3(x-2)(x-4) \ = \ 0.\)

\(\displaystyle Hence, \ (4,20) \ and \ (2,24) \ are \ critical \ points.\)

\(\displaystyle f \ " \ (x) \ = \ 6x-18 \ = \ 6(x-3) \ = \ 0, \ ergo, \ (3,22) \ is \ a \ point \ of \ inflection.\)

\(\displaystyle f \ " \ (4) \ > \ 0, \ hence \ relative \ min. \ and \ f \ " \ (2) \ < \ 0, \ relative \ max. \ (2nd \ derivative \ test).\)

\(\displaystyle See \ graph \ below.\)

[attachment=0:2dwqlsxy]old.jpg[/attachment:2dwqlsxy]

 

[Deckler]

Ah, I get it now. I think. Well enough to solve it mathematically anyway.