Question

Identify the extrema for each function. Classify each as a relative (local) or absolute (global) maximum or minimum value.

Answer 1

\[f(x) = 4x^3 - 12x^2 + 15\]

Answer 2

\[f \prime \left( x \right)=12x^2-24x\]
\[f \prime \left( x \right)=0~gives~12x^2-24x=0,12x(x-2)=0,x=0,2\]
\[f \prime \prime \left( x \right)=24x-12\]
\[f \prime \prime \left( 0 \right)=0-12=-12<0\]
There is a local maxima at x=0
f(0)=?
when x=2,
f"(2)=24*2-12=36>0
hence there is a relative minima and minimum value is
f(2)=?

Answer 3

correction

Answer 4

So I set the function equal to 0 and solve to find the local maximum?

Answer 5

\[f \prime \prime (x)=24x-24\]
then proceed further as i have done.

Answer 6

if you want local maxima then put x=0
if local minima then put x=2

Answer 7

f(0)=4*0^3-12*0^2-15=?
f(2)=?

Answer 8

Okay.

f(x)=4x3−12x2+15
f(0) = -15

-15 would be the local maximum? how do I know it's the "local" maximum and not the absolute maximum?

f(x)=4x3−12x2+15
f(2) = 32 - 48 + 15
f(2) = -1

The local (or absolute?) minimum equals -1.

Right?

Answer 9

Is the formula always f(2) = the local minimum, or just for this example?

Answer 10

sorry it is +15 not -15

Answer 11

Right.. :P

Answer 12

no here we found critical values by putting f"(x)=0
they are the values

Answer 13

?

Answer 14

we have to check with these critical values.

Answer 15

@Abbles are you a calculus student?

Answer 16

precalculus

Answer 17

but do you understand what @sshayer is doing with the derivatives?

Answer 18

Um... what are derivatives? :P

Answer 19

Also, what are "critical values"?

Answer 20

To be clear...

@sshayer was saying that to find the local maximum of a function, evaluate f(0) and to find the local minimum, evaluate f(2). Is that true for all functions? what about the absolute maximum/min?

Answer 21

@Abbles it sounds like you haven't learned derivatives yet since you're in pre-calculus. So I'd use a TI calculator (eg: TI83) or desmos or geogebra to graph the equation y = 4x^3 - 12x^2 + 15. Then use the min max feature to find the extrema.

https://www.desmos.com/calculator/1eheltseuy

The nice thing about desmos is that you simply click the extrema points to have the coordinates show up. See attached.

Answer 22

Alrighty, got my TI-84 ready. Is there no way to find the min/max without a calculator?

Answer 23

`Alrighty, got my TI-84 ready. Is there no way to find the min/max without a calculator?`
the only way without a calculator is to use calculus but that will come in a later class

anyways, graph the function using your TI-84 by typing in the equation in the y1 slot (after hitting the \(\Large y=\) key in the top left corner).

Then hit the \(\large \text{GRAPH}\) key at the top right corner to see the graph. Adjust the window so you see something similar to what I posted from the desmos graph. Once the graph is all set up, hit the blue \(\large \text{2ND}\) key and then hit the \(\large \text{TRACE}\) key (next to the GRAPH key).


Scroll down to 3:minimum and that will help find the local min. You'll have to provide the left boundary, the right boundary, and a guess somewhere in the middle. Repeat these steps, but do 4:maximum instead to get the local max. Again you'll need the left boundary, the right boundary, and a guess somewhere in the middle.

Answer 24

After I plugged in the function and clicked GRAPH, it's pulling up ERROR: INVALID DIMENSION

Answer 25

:/ any idea what's wrong? the calculator is fairly new and I haven't fooled around with it much.

Answer 26

you typed this in right? see attached

Answer 27

Yes pretty much... the model is actually a TI-84 plus ce... but same concept

Answer 28

what do you have for your window?

Answer 29

Xmin = -10 and Xmax = 10 ?

Not sure what all these mean...
Xscl = 1
Yscl = 1
Xres=1
etc.

Answer 30

this is what I have.

Answer 31

go ahead and type in what I have so we have the same window

Answer 32

Done. Still an error though. It says:

Check 1<dim(list)<999.
To set Plots0ff:
2nd STAT PLOT; Plots0ff
Check 1<dim(matrix)<99.
Check inverse of square matrix only.

Answer 33

I'm already taking one foreign language class right now lol.

Answer 34

so you have your stats plot on right now?

Answer 35

Sorry, what? :P I'm a complete newb when it comes to graphing calculators

Answer 36

when you hit the \(\Large y=\) key (top left corner), do you see Plot1, Plot2, or Plot3 highlighted at all?

Answer 37

if possible please show me a snapshot of the calculator screen you are looking at

Answer 38

Yes, Plot1

Answer 39

ok go up to that and turn that off

Answer 40

hitting enter over it should turn it off

Answer 41

Perfect, it's working now. So how do I find the mins/maxs of the graph? It's not showing any numbers on the graph for some reason...

Answer 42

hit the blue \(\large \text{2ND}\) key and then hit the \(\large \text{TRACE}\) key (next to the GRAPH key).

Scroll down to 3:minimum and that will help find the local min. You'll have to provide the left boundary, the right boundary, and a guess somewhere in the middle.

Answer 43

make sure that the left boundary is to the left of where you think the min is. You'll have to visually eyeball things a bit. Make sure the right boundary is to the right of the min

Answer 44

For the minimum I got X=2.0000015 and y=-1

Answer 45

due to some rounding error, you got 2.0000015 for x

I got 1.9999997 for x (see attached)

it should be x = 2 but the calculator only calculates to so many decimal digits and there's a bit of error.

Answer 46

Maximum: X=1.4772781 and Y-1.7075478

Answer 47

So the min is (2,-1)

Answer 48

Okay

Answer 49

`Maximum: X=1.4772781 and Y-1.7075478`
the local max is nowhere close to this

Answer 50

Let me try again

Answer 51

I should clarify that the local min is (2,-1)

it is not an absolute minimum

Answer 52

How do you determine if it's an absolute or local min/max?

Answer 53

Yep, for the max I got a perfect x = 0 and y = 15

Answer 54

If you have a certain interval, then the absolute max is the highest point on the interval. If you don't define an interval, then the absolute max is the entire domain. No intervals are stated in this problem, so we look at the entire cubic function. It goes on forever upward as x gets bigger, so there is NO absolute max at all

Answer 55

The same can be said in the other direction as well

Answer 56

so we have local extrema and no global (absolute) extrema

Answer 57

Got it

Answer 58

The absolute extrema can only be found if the domain is specified?

Answer 59

Mind going through one more with me? 3x^3 + 3x^2 - 30x + 24

Answer 60

Let's say you had this parabola

http://tutorial.math.lamar.edu/Classes/Alg/Parabolas_files/image005.gif

over the entire domain, the lowest point is at (1,2). This is the global/absolute min in this problem. Unless a different interval is stated, we use the domain as our interval

Answer 61

With that parabola, there is global max if no interval is stated because the graph keeps going up forever.

Answer 62

`Mind going through one more with me? 3x^3 + 3x^2 - 30x + 24`
sure, tell me how far you get with this next one

Answer 63

Okay, for this one I just need the local max. x=-2 and y=72 sound about right?

Answer 64

Is it asking to round to the nearest whole number?

Answer 65

No...

Answer 66

To the nearest thousandth

Answer 67

This is what I get

https://www.desmos.com/calculator/pwvx206d96

see attached

Answer 68

72.5 seem closer? how am I supposed to find an answer as accurate as the thousandth place?

Answer 69

72.5 is what your TI84 is saying?

Answer 70

No, from looking at your graph... that's what it looks to be

Answer 71

oh, I recommend you use your TI84 again or use desmos like I did. Look at the last attachment I posted

Answer 72

I got it wrong :/ I put 72.5; the correct answer was 72.577-72.578

Oh well. At least now I know how to do it. Thanks for your help :)

Answer 73

it should be 72.578

hopefully you see the coordinates on that attachment?

Answer 74

this one here

http://assets.openstudy.com/updates/attachments/57688b3de4b0144f08365531-jim_thompson5910-1466474647259-screenshot_6.png

Answer 75

Thanks. I don't get why my calculator was being so inaccurate :/ oh well..