What is the local maximum value of the function? (Round answer to the nearest thousandth.)

g(x) = 3x^3 + 3x^2 - 30x + 24

Question

What is the local maximum value of the function? (Round answer to the nearest thousandth.)

g(x) = 3x^3 + 3x^2 - 30x + 24

superdavesuper · Answer

u need to know differentiation to solve this: at local max, g'(x)=0. solve for x then u will find the value of g(x) there.

mathmale · Answer

supernerd:
Please find the derivative of the given function g(x) and share your result here.

anonymous · Answer

For the derivative I got: 9x^2+6x-30 @mathmale

mathmale · Answer

Great.  Now, as Dave has already suggested, please set that = to 0 and solve for x.

You may find it easier to factor out the common factor 3 before you do this.

mathmale · Answer

Hint:  the results of using the quadratic formula in this case to find roots are not pretty!

superdavesuper · Answer

thanks @mathmale

@supernerd

it may not be pretty but g'(x)=0 is solvable; here is how g(x) look like :)

anonymous · Answer

g(x)=

anonymous · Answer

g'(x)=9x^2+6x-30
         9x^2+6x-30=0
critical vaule= x=-(sqrt(31)+1)/3 or x=(sqrt(31)-1)/3

anonymous · Answer

I simplified those all further and got 1.5226, is that correct?

mathmale · Answer

How could you check this result for yourself?  Doing so would make you a supernerd.
Hint:  Graph the given function and then vary x until you get x= 1.523.  See a max or a min?  Of course there are other ways to do this; you could use synth div, but that would be messy.

superdavesuper · Answer

@mathmale is right

if u look at the attached graph to my previous post, u can see that the max is quite different. so i think u made a mistake somewhere.

superdavesuper · Answer

oh wait @supernerd i think u found the local min instead of the max...try the other root for g'(x)=0 plz

mathmale · Answer

Right, Dave; I did a quick calculator calculation of the value of y at supernerd's x=1.523 and found that the resulting point, (1.523,-4.133), is indeed a relative minimum of the given function.

superdavesuper · Answer

hehehe i cheated....i used the web site that begins w "w" n thats where i got the graph too ;)

mathmale · Answer

I think that's a fine approach; we know how to graph manually, so if we can do it much more quickly online, why not?

anonymous · Answer

Ah, I keep trying but I just keep getting the same answer, how do we find the maximum as opposed to the minimum?

superdavesuper · Answer

there should be two roots to g'(x)=0 as it is a quadratic eqn. 1 is the max n the other is the min. So try the other root plz :)

anonymous · Answer

I got 2.1893! Is that correct? @superdavesuper

mathmale · Answer

Not to be unkind, but I think you should be thinking about how y ou could check your own answers. See Dave's illustration: http://assets.openstudy.com/updates/attachments/52e404f8e4b096fa40598f39-superdavesuper-1390677131242-untitled.png and ask yourself whether or not this graph seems to have a local max when x=2.1893.

mathmale · Answer

"critical value= x=-(sqrt(31)+1)/3 or x=(sqrt(31)-1)/3"   came from "annasjuice."

Try evaluating these again.  One value should be 1.523; the other one should be (-).

anonymous · Answer

Oh! This time I got: -2.1893

mathmale · Answer

A lot better!  Now how are you going to determine whether or not that's "right"?

anonymous · Answer

plug it in to the original equation? Or graph it?

mathmale · Answer

Look at Dave's original illustration. Does it appear that his graph has a max at your -2.189? http://assets.openstudy.com/updates/attachments/52e404f8e4b096fa40598f39-superdavesuper-1390677131242-untitled.png

anonymous · Answer

Yes!