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find the critical numbers f(x)=-x^4+4x^3+5

Mathematics
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As they always say: find \(f'(x)\) and equate that to \(0\).
@chochko What would \(f'(x)\) be here?
-4x^3+12x^2

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Other answers:

ANd then?
Well that's correct - now equate it to zero.
Solve this equation:\[\rm 4x^3 + 12x^2 = 0\]
-4x^3+12x^2=0
Yeah, I meant -4x^3 + 12x^2 = 0
sry my computer is really slow. Would you factor out a 4?
Yu.. 4x^2 ( x-3) = 0 4x^2 = 0 x = 0 x-3 = 0 x = 3
Mine too, and you can factor 4x^2 out.
Yep, and what Yahoo! said are the critical points. ;)
thank you but im not done . that is just one part..can you help yahoo or ParthKohli?
Sure.
the next part is to differentiate it again :P to determine its nature
You guys must be in the same school.
lol
using the second derivitive test which is -12x^2+24x=0 set to 0 to determine all relative extrema, indicating the x and Y values and wheteher is a max or a min. Help please
lol @ParthKohli: the next step is common sense. @chochko: with the differentiataed function sub the \(x\) that you found in the first step if its a negative then its concave down => max. if its concave up then => min
Sorry, was on another question, and yeah - that's the way I was taught too.
so in -12x^2+24x Sub the x for the critical numbers and the result is the relative extrema? is that the Max and Min as well?
max or min concave up => min concave down => max
so which part is the relative extrema? i really suck at math btw?
well i googled; relative extrema and max/ min extrema is the same thing
f''(0)=0 min f''(3)=-36 Max Is this correct?
i dont know; i dont have a calculator with me but if its you have to check if it has a horizontal point of inflexion
Ok, so the intervals are the critical points right? where are they increasing and decreasing?
critical points are the stationary points on the curve since \(f(x)=0\) you differentiate it twice to determine if its increasing or decreasing increasing \(f''(x) >0\)
i think i made a mistake; increasing \(f'(x)>0\)
when the function is decreasing \(f'(x)<0\)
i mean you differetiate it once; to see if its increasing or decreasing sorry; im sick
so using the f'(x)=0 and plug in 0 and 3 for x to find where f(x) is increasing or decreasing?
oh no no im getting tired @ParthKohli: may like to explain :)
ok thanks a lot
@ParthKohli are you able to help me finish this one
I think I should explain; since Parth is hiding :P \(f'(x) = 0\) you are finding the \(x\) values that is the STATIONARY point. \(f'(x) >0\) is the \(x\) values were the function is INCREASING \(f'(x)<0\) is where the function is DECREASING. The second derivative is the deirivative of the derivative. The curve \(y=f(x)\) is concave up when \(f''(x)>0\) and concve down when \(f''(x)<0\)
Sry, back.
Didn't notice the tag right there. :P
Increasing on ( -inf, 0) (0,3) then decreasing (3, inf) idk if its correct?
Well your \(\rm f'(x)\) function is \(\rm -4x^2 +12x \) and the function is stationary at x = 3,0.
What is the question again?
State the intervas for which f(x) is increasing or decreasing So i tested the values -1, 1, 4 And got F'(-1)=16 inc f'(2)=8 inc f'(4)=-64 dec. idk if its correct?
That's right.
So it starts decreasing after 3.
(-inf, 0) (0,3) increasing then (3, inf) decreasing Are the intervals right?
Let's test f'(-5). f'(x) = -4x^2 + 12x f'(-5) = -4(-5)^2 + 12(-5) = -4(25) - 60 = -100 - 60 = -160
It's decreasing at f(-5).
The best way is to ask Wolfram for the graph. :)
huh? whos that?
http://wolframalpha.com
Enter your function and see its graph.
ok can you help me find the possible inflection points
@ParthKohli can you help just a lil bit left
Sorry, back.
Damn... I've disabled the tag notifs.

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