
But it was only fantasy
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Currently studying for my Calculus exam on Monday but am stuck on some particular problem.
I took the first derivative and got:
12x^2 + 30x + 18
Then I factored it:
6(2x^2 + 5x + 3)
Used the quadratic formula and got x=1 and x=1.5
My question is... what do I do after that? srs
Thanks in advance.
(I know there's an official math thread but I would literally have to wait a few days before getting an answer)
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you know the xvalues of the extrema so plug them back into the original function and get your max/min values and see which are absolute

Spirit Crusher
Plug your critical numbers into the original equation.
First find f'(0).
I didn't check your work but that would make 1 and 1.5 your critical numbers.
f(1)
f(1.5)
f(0)
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Registered User
f'(x) = 12x^2 + 30x+18
zeros = .5,3
plug em back into f(x)

Registered User
The next step is to test the extremes you got by setting the derivative to = 0.
just plug in a slightly larger/smaller value for x and see what you get in the function. this way you know if they are max's or mins
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Omnomnomnom'ing
Originally Posted by cobhc
Plug your critical numbers into the original equation.
First find f'(0).
I didn't check your work but that would make 1 and 1.5 your critical numbers.
f(1)
f(1.5)
f(0)
Also check your boundary conditions.
and FYI, the derivative is 12x^2 + 30x + 18
Double check your answers by plugging it into your calculator and using the max,min thing (TI83)

Registered User
use critical numbers including 2 and 5 and plug them back into f(x) not f'(x)

But it was only fantasy
Originally Posted by cobhc
Plug your critical numbers into the original equation.
First find f'(0).
I didn't check your work but that would make 1 and 1.5 your critical numbers.
f(1)
f(1.5)
f(0)
I did EXACTLY this, but the answer key from my professor shows completely different numbers... this is why I came here.
This is what he wrote down.
Am I missing a step after that?
EDIT: Wtf how are you guys getting 5 and 3? :[
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set the derivative equal to 0. solve for x. plug x back into original equation.
edit: your derivative is completely wrong. da fug mayne.
18 + 30x  12x^2
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Hamiltonian operator
Originally Posted by SCARH
Currently studying for my Calculus exam on Monday but am stuck on some particular problem.
I took the first derivative and got:
12x^2 + 30x + 18
Then I factored it:
6(2x^2 + 5x + 3)
Used the quadratic formula and got x=1 and x=1.5
My question is... what do I do after that? srs
Thanks in advance.
(I know there's an official math thread but I would literally have to wait a few days before getting an answer)
to distinguish between max and min take 2nd derivative. plug in your calculated value. if 2nd derivative is negative, its a max. if 2nd derivative is positive, its a min

Spirit Crusher
Originally Posted by SCARH
I did EXACTLY this, but the answer key from my professor shows completely different numbers... this is why I came here.
This is what he wrote down.
Am I missing a step after that?
Well for one, the roots of f'(x) are x = 3 and x = .5
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But it was only fantasy
Originally Posted by StrongScience62
set the derivative equal to 0. solve for x. plug x back into original equation.
edit: your derivative is completely wrong. da fug mayne.
18 + 30x  12x^2
Not sure if srs, you just typed the same thing. But thanks
EDIT: Oh, I see that I ignored the negative sign on the original equation when I rearranged it.
Originally Posted by 4ea
to distinguish between max and min take 2nd derivative. plug in your calculated value. if 2nd derivative is negative, its a max. if 2nd derivative is positive, its a min
Alright, I will do that.
Last edited by SCARH; 05032012 at 11:15 PM.
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Registered User
1. your first derivative is wrong, its f'(x)=12x^2+30x18
which factors to 6(x3)(2x+1)
so critical numbers are, 3, 1/2 and the interval they gave u .
so plug 3,/2 and the end points of the interval into f(x)

itsmenara
Originally Posted by moneymango
f'(x) = 12x^2 + 30x+18
zeros = .5,3
plug em back into f(x)
this is what i got. just plug 1/2 and 3 into the original equation.
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Originally Posted by SCARH
Not sure if srs, you just typed the same thing. But thanks
hey retard. negative signs matter. check again.
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Hamiltonian operator
Originally Posted by SCARH
cool. i think you got the process down for getting the critical values. the second step with the 2nd derivatives is the final step for this problem

Spirit Crusher
Also, don't forget that your intervals are critical numbers.
f(2)
f(5)
f(0)
f(3)
f(.5)
And voila.
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But it was only fantasy
Originally Posted by gamblingman
1. your first derivative is wrong, its f'(x)=12x^2+30x18
which factors to 6(x3)(2x+1)
so critical numbers are, 3, 1/2 and the interval they gave u .
so plug 3,/2 and the end points of the interval into f(x)
The main problem with these questions is that I suck ass at factoring. I tried using the quadratic formula... lol
What do I look out for when factoring equations like these?
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Hamiltonian operator
Originally Posted by SCARH
The main problem with these questions is that I suck ass at factoring. I tried using the quadratic formula... lol
What do I look out for when factoring equations like these?
quadratic formula will always work. id say glance at the expression and see if the factoring jumps out at you, if it doesnt, just grind it out with quadratic formula

Spirit Crusher
Originally Posted by SCARH
The main problem with these questions is that I suck ass at factoring. I tried using the quadratic formula... lol
What do I look out for when factoring equations like these?
Well your derivative before you simplify will be: 12x^2 + 30x + 18
Pull a 6 out to make life easier.
6(2x^2 + 30x + 18)
Factor out further.
6(2x + 1)(x3)
2x + 1 = 0
x = 1/2

x3 = 0
x = 3
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But it was only fantasy
Originally Posted by cobhc
Well your derivative before you simplify will be: 12x^2 + 30x + 18
Pull a 6 out to make life easier.
6(2x^2 + 30x + 18)
Factor out further.
6(2x + 1)(x3)
2x + 1 = 0
x = 1/2

x3 = 0
x = 3
It looks so simple once you have it all figured out. That's the only thing that keeps me going. Thanks
EDIT: I have a question from a more recent test. I'm going to try it out and post it here to see if I get it right. Won't look at the answer key just yet.
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Spirit Crusher
Originally Posted by SCARH
It looks so simple once you have it all figured out. That's the only thing that keeps me going. Thanks
Remember the formula: ax^2 + bx + c = (dx + e)(fx + g)
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But it was only fantasy
Alright, here we go.
This is from the '11 test.
f'(x)= 6x^2  18x  24
6(x^2  3x  4)
Quadratic formula
x=4;x=1
x=4
x=1
x=2
x=7
Plug all of these numbers into the original equation and I get
x@2= 11
x@1= 28
x@4= 97
x@7= 92
Abs max = f(7)
Abs min = f(4)
Does that look correct?
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Spirit Crusher
f'(x) = 6x^2  18x  24
6(x^2 3x 4)
6(x4)(x+1)
x=4
x=1
f(4)
f(1)
f(0)
f(2)
f(7)
I didn't plug them in but I think you're good to go OP. Good job.
Just a something to remember for the exam, incase your professor tries to phuck with you:
If any of your roots are not in the intervals given, then you don't have to plug them in.
So if you had an x = 8 in this case, you would just ignore it.
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derive, crit numbers, put on number line, check if positive or negative, done
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But it was only fantasy
Originally Posted by cobhc
f'(x) = 6x^2  18x  24
6(x^2 3x 4)
6(x4)(x+1)
x=4
x=1
f(4)
f(1)
f(0)
f(2)
f(7)
I didn't plug them in but I think you're good to go OP. Good job.
Just a something to remember for the exam, incase your professor tries to phuck with you:
If any of your roots are not in the intervals given, then you don't have to plug them in.
So if you had an x = 8 in this case, you would just ignore it.
Far from it, but thanks! It's covering all Chapter 4.
L'Hospital
Curve Sketching
Optimization
and a few other things.
I'll do optimization tomorrow, since curve sketching seems a lot easier now.
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Spirit Crusher
Originally Posted by SCARH
Far from it, but thanks! It's covering all Chapter 4.
L'Hospital
Curve Sketching
Optimization
and a few other things.
I'll do optimization tomorrow, since curve sketching seems a lot easier now.
Feel free to pm with questions brah
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