Finding Relative Maximum and Minimum

• July 5th 2009, 07:14 PM
mvho
Finding Relative Maximum and Minimum
Q1. Test the Function $g(x)=6x^3-(108)x^2+(630)x-2$ for relative maximum and minimum. Find the critical values. Use the 2nd derivative test if possible.

My solution:

$g'(x)=18x^2-216x+630$ if follows that the critical values of g are x1= -7 and x2=-5.

*
To find the critical numbers, I set g'(x) =0. This is where I think I went wrong.

First, I factored out a "6"

$6(3x^2-36x+105)$
Then, I factored out a 3 and that's how I arrive on my answer of -7 and -5.

Q.2 (Related to Q1)

Since $g"(x)=36x-216$ <-- My guess, we have that g"(x1)= ? greater or less than 0 and g"(x2) = ? greater or less than 0. (I need to find out values for x1 and x2)

Therefore, by the 2nd derivative test, there is a either a relative max or min when x=x1 and there is a relative max or min when x=x2. (Choose either or)

TIA
• July 5th 2009, 07:18 PM
VonNemo19
Quote:

Originally Posted by mvho
Q1. Test the Function $g(x)=6x^3-(108)x^2+(630)x-2$ for relative maximum and minimum. Find the critical values. Use the 2nd derivative test if possible.

My solution:

$g'(x)=18x^2-216x+630$ if follows that the critical values of g are x1= -7 and x2=-5.

* To find the critical numbers, I set g'(x) =0. This is where I think I went wrong.

First, I factored out a "6"

$6(3x^2-36x+105)$
Then, I factored out a 3 and that's how I arrive on my answer of -7 and -5.

Q.2 (Related to Q1)

Since $g"(x)=36x-216$ <-- My guess, we have that g"(x1)= ? greater or less than 0 and g"(x2) = ? greater or less than 0. (I need to find out values for x1 and x2)

Therefore, by the 2nd derivative test, there is a either a relative max or min when x=x1 and there is a relative max or min when x=x2. (Choose either or)

TIA

What exactly did you do after you factored out the three? You were on the right track.

1. take the derivative.
2. Set it to zero
3. the solution set are your critical numbers.
• July 5th 2009, 07:25 PM
mvho
I am such an Idiot!! The critical values are 5 and 7. I originally had it factored down to $(x-7)(x-5)$

x-7=0
x=7

DUH!

I am a bit confused on the 2nd part though. I know my 2nd derivative is correct:

$g"(x)=36x-216$ but am unsure about this whole " we have that g"(x1)=_____ greater OR less than 0 and g"(x2)=________ greater or less than 0.

Therefore, by the second derivative test there is a relative max or min when x=x1 and there is a relative max or min when x=x2.

Can someone explain that in Simpler terms?
• July 5th 2009, 07:26 PM
McScruffy
Once you've found your critical numbers (by setting the first derivative to zero and solving for x), you can apply the second derivative test by putting those critical values into the second derivative. If the value is positive then the function at the given point is a relative minimum, and vise versa for a negative value returned from the second derivative test. Hope that helps.
• July 5th 2009, 07:30 PM
VonNemo19
Quote:

Originally Posted by mvho
I am a bit confused on the 2nd part though. I know my 2nd derivative is correct:

$g"(x)=36x-216$ but am unsure about this whole " we have that g"(x1)=_____ greater OR less than 0 and g"(x2)=________ greater or less than 0.

Therefore, by the second derivative test there is a relative max or min when x=x1 and there is a relative max or min when x=x2.

Can someone explain that in Simpler terms?

I had the same question recently. Check this out.

http://www.mathhelpforum.com/math-he...tive-test.html
• July 5th 2009, 07:41 PM
mvho
Quote:

Once you've found your critical numbers (by setting the first derivative to zero and solving for x), you can apply the second derivative test by putting those critical values into the second derivative. If the value is positive then the function at the given point is a relative minimum, and vise versa for a negative value returned from the second derivative test. Hope that helps.
Ok, that is much easier to understand.

So, since my critical numbers were 5 and 7 in this case, I simply plug them into my g"(x)=36x-216.

When I plug in "5", the value is negative so I have a maximum

When I plug in "7", the value is positive so I have a minimum.
• July 5th 2009, 07:45 PM
McScruffy
Exactly right.