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

- anonymous

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

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- ParthKohli

As they always say: find \(f'(x)\) and equate that to \(0\).

- ParthKohli

@chochko What would \(f'(x)\) be here?

- anonymous

-4x^3+12x^2

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## More answers

- anonymous

ANd then?

- ParthKohli

Well that's correct - now equate it to zero.

- ParthKohli

Solve this equation:\[\rm 4x^3 + 12x^2 = 0\]

- anonymous

-4x^3+12x^2=0

- ParthKohli

Yeah, I meant -4x^3 + 12x^2 = 0

- anonymous

sry my computer is really slow. Would you factor out a 4?

- anonymous

Yu..
4x^2 ( x-3) = 0
4x^2 = 0
x = 0
x-3 = 0
x = 3

- ParthKohli

Mine too, and you can factor 4x^2 out.

- ParthKohli

Yep, and what Yahoo! said are the critical points. ;)

- anonymous

thank you but im not done . that is just one part..can you help yahoo or ParthKohli?

- ParthKohli

Sure.

- Mimi_x3

the next part is to differentiate it again :P
to determine its nature

- ParthKohli

You guys must be in the same school.

- ParthKohli

lol

- anonymous

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

- Mimi_x3

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

- ParthKohli

Sorry, was on another question, and yeah - that's the way I was taught too.

- anonymous

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?

- Mimi_x3

max or min
concave up => min
concave down => max

- anonymous

so which part is the relative extrema? i really suck at math btw?

- Mimi_x3

well i googled; relative extrema and max/ min extrema is the same thing

- anonymous

f''(0)=0 min
f''(3)=-36 Max Is this correct?

- Mimi_x3

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

- anonymous

Ok, so the intervals are the critical points right? where are they increasing and decreasing?

- Mimi_x3

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 \(f''(x) >0\)

- Mimi_x3

i think i made a mistake; increasing \(f'(x)>0\)

- Mimi_x3

when the function is decreasing \(f'(x)<0\)

- Mimi_x3

i mean you differetiate it once; to see if its increasing or decreasing
sorry; im sick

- anonymous

so using the f'(x)=0 and plug in 0 and 3 for x to find where f(x) is increasing or decreasing?

- Mimi_x3

oh no no
im getting tired @ParthKohli: may like to explain :)

- anonymous

ok thanks a lot

- anonymous

@ParthKohli are you able to help me finish this one

- Mimi_x3

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 \(f''(x)<0\)

- ParthKohli

Sry, back.

- ParthKohli

Didn't notice the tag right there. :P

- anonymous

Increasing on ( -inf, 0) (0,3) then decreasing (3, inf)
idk if its correct?

- ParthKohli

Well your \(\rm f'(x)\) function is \(\rm -4x^2 +12x \) and the function is stationary at x = 3,0.

- ParthKohli

What is the question again?

- anonymous

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?

- ParthKohli

That's right.

- ParthKohli

So it starts decreasing after 3.

- anonymous

(-inf, 0) (0,3) increasing then (3, inf) decreasing
Are the intervals right?

- ParthKohli

Let's test f'(-5).
f'(x) = -4x^2 + 12x
f'(-5) = -4(-5)^2 + 12(-5)
= -4(25) - 60
= -100 - 60 = -160

- ParthKohli

It's decreasing at f(-5).

- ParthKohli

The best way is to ask Wolfram for the graph. :)

- anonymous

huh? whos that?

- ParthKohli

http://wolframalpha.com

- ParthKohli

Enter your function and see its graph.

- anonymous

ok can you help me find the possible inflection points

- anonymous

@ParthKohli can you help just a lil bit left

- ParthKohli

Sorry, back.

- ParthKohli

Damn... I've disabled the tag notifs.

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