Question

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

Answer 1

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

Answer 2

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

Answer 3

-4x^3+12x^2

Answer 4

ANd then?

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

Yu..

4x^2 ( x-3) = 0

4x^2 = 0

x = 0


x-3 = 0

x = 3

Answer 11

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

Answer 12

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

Answer 13

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

Answer 14

Sure.

Answer 15

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

Answer 16

You guys must be in the same school.

Answer 17

lol

Answer 18

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

Answer 19

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

Answer 20

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

Answer 21

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?

Answer 22

max or min
concave up => min

concave down => max

Answer 23

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

Answer 24

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

Answer 25

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

Answer 26

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

Answer 27

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

Answer 28

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

Answer 29

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

Answer 30

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

Answer 31

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

sorry; im sick

Answer 32

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

Answer 33

oh no no

im getting tired @ParthKohli: may like to explain :)

Answer 34

ok thanks a lot

Answer 35

@ParthKohli are you able to help me finish this one

Answer 36

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

Answer 37

Sry, back.

Answer 38

Didn't notice the tag right there. :P

Answer 39

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

Answer 40

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

Answer 41

What is the question again?

Answer 42

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?

Answer 43

That's right.

Answer 44

So it starts decreasing after 3.

Answer 45

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

Answer 46

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

Answer 47

It's decreasing at f(-5).

Answer 48

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

Answer 49

huh? whos that?

Answer 50

http://wolframalpha.com

Answer 51

Enter your function and see its graph.

Answer 52

ok can you help me find the possible inflection points

Answer 53

@ParthKohli can you help just a lil bit left

Answer 54

Sorry, back.

Answer 55

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