Question

math section down someone help

f(x)= x^3-2x^2

relative min and max using second derivative test?

Answer 1

You only need to post your question once. The math section is not down.

Answer 2

im sorry can you link me to it? it wasnt work for me.

Answer 3

Aren't you already in the math section bro?

Answer 4

if you know how to do this, can you help me?
plsss

Answer 5

they split them into subsections? before algebra,calc,etc were all in one sections

Answer 6

pls can you help me wi

Answer 7

*with this

Answer 8

im preparing for a test plssss

Answer 9

1. Take the derivative
2. Set it to zero, solve for x
3. Take second derivative
4. Input value of x
5. If negative, then x is max, If positive, then x is min

Answer 10

ok so far ive got
f`(x)=3x^2-4x
3x^2-4x=0
x(3x-4)=0

whats next?

Answer 11

Take the derivative of f'(x) = 3x^2 - 4x to find f''(x)

To find the zeroes of f'(x) use the zero product property

Answer 12

im confused...can you demonstrate it pls?

Answer 13

Pretend that f'(x) is your new h(x)

Now take the derivative of h(x) = 3x^2 - 4x

Answer 14

6x-4 f"(x)

Answer 15

??

Answer 16

Okay good

Answer 17

We need to input the zeroes of f'(x) into f''(x)

Answer 18

I'm pretty sure you know how to solve

x(3x - 4) = 0

Answer 19

x=0, x=4/3
??

Answer 20

exactly, now put those values into f''(x) one at a time.

Answer 21

6(0)-4= -4
6(4/3)-4= 4

Answer 22

??

Answer 23

Okay, so which is min and which is max?

Answer 24

min=-4 max=4

Answer 25

No, you have it backwards. Re-read the steps I presented to you.

Answer 26

oh max=-4 min=4

Answer 27

thanks i have another one f(x)=x^3 + 3x^2 but let me attempt it first

Answer 28

That's the only thing about second derivative. You have to remember that if f''(x) produces negative value, then f(x) is max, but if f''(x) produces positive value, then f(x) is min

Answer 29

Yes, you should try it first.

Answer 30

No reason why you can't finish it. The only thing is you need to find f(0) and f(4/3) for the previous problem.

Answer 31

ive got...

Answer 32

If I were you, I would make a chart to keep track of everything.

Answer 33

f'(x)=3x^2+6x=0
3x(x+2)=0
x=0
x=2
f"(x)=6x + 6)
6(0)+6=6
6(2)+6=18

Answer 34

You mean x = -2

Answer 35

how so

Answer 36

Take a moment to think about it

Answer 37

hey the program went off are you there

Answer 38

i figured out

Answer 39

Okay good

Answer 40

i have a quite similar question...now i have to find the absolute max and min of f(x)= x^3 + 3x^2 on the interval [1,1]...whats the difference between relative and absolute anyhow

Answer 41

For the first couple problems, there was only one min and one max. Now for this, you will have to choose the highest min and max between several values.

Answer 42

To do this, you do the same exact thing you did before, except this time, you will also have to test x = 1 and x = -1

Answer 43

so i take the first and second derivatives then test -1 and 1? into which derivative?

Answer 44

Basically, do everything you did before (all five steps while ignoring x = -1,1 for the moment).

Then after you have completed the five steps, then also evaluate x = -1,1

Answer 45

might need a demonstration...its kind of confusing..let me attempt first

Answer 46

Basically just pretend you don't have [1,1] for now.

Answer 47

Pretend that you have to find the relative min and max of f(x)= x^3 + 3x^2

Answer 48

So perform the five steps as usual.

Answer 49

Let me know what you get afterwards

Answer 50

f'(x)=3x^2+6x
x(3x-6)=0
x=0
x=6/3

f"(x)=6x+6
6(0)+6=6
6(6/3)+6=

Answer 51

??

Answer 52

6/3 = 2 btw

Answer 53

x^3 + 3x^2 was orginal function

Answer 54

yea i forgot to type 18

Answer 55

What? Where does 18 come from? I said 6/3 = 2

Answer 56

2 * 6 + 6=18?

Answer 57

Okay, but

Answer 58

We only need to consider x values between -1 and 1 so x = 2 is out

Answer 59

ohhh ok

Answer 60

So basically you need to test:

f''(-1)
f''(0)
f''(1)

Answer 61

ohh gotcha

Answer 62

If I were you, I would just simply test

f(-1)
f(0)
f(1)

That will tell you the min and max

Answer 63

You're only looking for the extreme min and extreme max

Answer 64

i see thanks for the help. i have several more questions though

Answer 65

I bet you do

Answer 66

how do i solve
3^(x^2 - x) = (1/9)^(3x)

Answer 67

im preparing for a test sorry

Answer 68

For this, I'll need to write out the steps so that you can see them

Answer 69

thats fine

Answer 70

can you pls?

Answer 71

Hang on a sec bro

Answer 72

ok

Answer 73

Okay, I have it

Answer 74

I'll send you a link to vyew for this

Answer 75

ok no problem

Answer 76

is that the final answer

Answer 77

No, bro. It is freezing up

Answer 78

ohhh ok

Answer 79

Good luck figuring out the steps

Answer 80

You probably got lost somewhere. I just know it

Answer 81

yeah bro ill try...you think you can help with this last one? 3+ 2e^(-t)= 7

Answer 82

That's an easy one bro

Answer 83

have to solve the equation....ill try to go over it and ask my peers for help

Answer 84

damn bro im so tire from all this studyy...i really cant figure out that one

Answer 85

Go back to vyew bro

Answer 86

ok

Answer 87

thanks for all the help bro