Question

f(x)= 1/3 x^3+1/2 x ^2 x^2-6x+1
the function is only valid in the domain -5 bigger or qual = x = smaller or equal 5
how do you find the maxima and the minima?

Answer 1

@wilbert06

Do you know how to find critical points?

Answer 2

reminder: derivatives

Answer 3

no clue

Answer 4

i know how to derrive but i dont understand the question

Answer 5

This is actually not that hard, I'll help. First, take your derivative. Find f'(x). Type it in here when you've got it, ok?

Answer 6

x^2 +x -6x^-1+ 0

Answer 7

f'(x) =x^2 +x -6x^-1+ 0

Answer 8

@agentx5 there you have it what do i do next?

Answer 9

Ah so not
\[\frac{d}{dx}(x^3/3+(x^2 x^2)/2-6 x+1) = 2 x^3+x^2-6\]

You had this as your original then?
\[\int\limits (x^2+x-6/x+0) dx = 1/6 x^2 (2 x+3)-6 log(x)+constant\]
(integral goes in reverse)

Answer 10

I don't think I have your original function quite understood. Show me with parenthesis what you meant.

Answer 11

ok let me try to write it doen better

Answer 12

The next step will be to evaluate for 0's, because that's when the SLOPE is 0, or it's maxed or minimized out. Make sense? It's got no slope at the tip of a peak or valley.

Answer 13

\[f(x)= 1/3 x ^{3}+ 1/2^{2}-6x+1\]

Answer 14

that was the given fucntion

Answer 15

so step one would be differentiate it?

Answer 16

\[1/2^2+1=\frac{1}{4}+1=\frac{5}{4}\] So that's just your constant, yeah?

And yes, step one is exactly that

Answer 17

so after its differentiated you solve x, that means you make the equation = 0?

Answer 18

Check to see if you have this:
\[ \frac{d}{dx}(\frac{x^3}{3}-6 x+\frac{5}{4}) = x^2-6 \]

Answer 19

Then you'll have your f'(x) function. The critical points are f'(0), plug in zero. As you might be able to see because it's a linear function, you're only going to have one point.

Answer 20

yes its that

Answer 21

ahhh i see the zero is when the slope is a straight line

Answer 22

Take the derivative again of x^2-6 and see if it's +, -, or 0. This tells you the concavity. It's just 2, which is a + number so it opens upward there.

Answer 23

Yeah you'll eventually be able to do this in your head, and in Calc2 or Calc3 you'll really need to in fact.

Answer 24

Here let me give you the graph and you'll be facepalming :D

Answer 25

ok

Answer 26

See if while I'm grabbing this you can find coordinate for the original function for that point when the slope is zero. There's more than one because it's an x cubed function originally

Answer 27

It's going to cross the x axis in 3 locations, 3 roots

Answer 28

ok so again, step 1 would be differentiate, step 2 would be solve x for zero and then?

Answer 29

the original would be the integral?

Answer 30

derivative and integral are inverse operations

just like...
addition and subtraction
multiplication and division
exponent and logarithm

And here's your graph to look at:

Answer 31

Forward & reverse, basically. This is a core concept of ALL mathematics. If you can do it one way, you better be able to reverse it with the right conditions.

Answer 32

Technically it's derivative and anti-derivative and the anti-derivative with the conditions make it into the operation called integral

Answer 33

ok aha

Answer 34

What you're doing to find critical points is finding when slope is zero. And in this case the GLOBAL minimum and maximum are -∞ and ∞

Answer 35

You were asked to find local I presume, which is those peaks or valleys "bends" when it changes direction.

Answer 36

ohh ok local and global theres a difference

Answer 37

Not always. But for this kind of function? yes.

Answer 38

yeah i guess the local you know between 2 and 4 in think

Answer 39

they did ask between -5 and 5

Answer 40

Between 5 and -5 is your domain, what they're restricting you to. Major hint there it's local min's or max's in that section

Answer 41

Graph the parabola y=x^2-6 and you'll find where it crosses the x-axis (when (f'(x)=0) is EXACTLY where the slope is zero

Answer 42

abc formula not aply able?

Answer 43

square root of 6?

Answer 44

ax^2+bx+c? Um, no that's only for special cases of polynomials

Answer 45

Yes, it's the squareroot of 6, plus or minus

Answer 46

The local max is at the minus, the local min is at the plus

Answer 47

ohhh !

Answer 48

and wala that was it?

Answer 49

Yep!

Answer 50

Don't forget to specific coordinates though

Answer 51

Got to answer the original problem, but now you know what x has to be in order for slope to be zero for the original problems and which is which

Answer 52

so the answer would be ,step 1) differentiate
step 2) find y=0
step 3) is the answer?

Answer 53

y'=0 , there's a diff ;-)

Answer 54

haha ok

Answer 55

Not trying to be nitpicky but it's kinda important to note it's the derivative

Answer 56

yes yes

Answer 57

(x , y )

Answer 58

ah sooo the specific coordinates

Answer 59

i would have to fill in the square root of 6 in th original formula?

Answer 60

@agentx5 is that so?

Answer 61

@agentx5 thanks for the help! i see it now atleast i think i understand the method

Answer 62

@agentx5 goodnight