local minimum of f(x) = x^3 - 9x^2 - 120x + 6

Math wiz ya rite 09

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For what value of x does the function f(x) = x^3 - 9x^2 - 120x + 6 have a local minimum?

a) 10
b) 4
c) 3
d) -4
e) -10
 
Re: local min

Please show us your work/thoughts - so that we know where to begin to help you.
 
Math wiz ya rite 09 said:
i think that the first step is to tak 1st derivative... then what do i do?
Apply the First Derivative Test, like they showed in class (and you can review it in your textbook, if you're fuzzy on it) to find the critical points. If you've learned the Second Derivative Test, apply that, too. Otherwise, find the sign of the derivative on either side of the critical points to determine max/min points.

Eliz.
 
First of all you are dealing with a polynominal whose domain is -infinity to infinity.

Hence no calculus is necessary. Just plug in the values (10,4,3,-4,-10) and see which one gives you a local minimum.
 
Dr. Flim-Flam said:
...no calculus is necessary. Just plug in the values (10,4,3,-4,-10) and see which one gives you a local minimum.
How will this method help when the question isn't multiple-choice? :shock:

Also, the lowest (or highest) value (from plugging the answer options into the polynomial) won't always be a local minimum (or maximum). :oops:

Eliz.
 
Dr. Flim-Flam said:
First of all you are dealing with a polynominal whose domain is -infinity to infinity.

Hence no calculus is necessary. Just plug in the values (10,4,3,-4,-10) and see which one gives you a local minimum.

I reluctantly agree - since "none of the above" is not a choice (guaranteed answer in one of those points). But I think this method is very computation intensive and can lead to mistakes (without the use of a spreadsheet like Xcel).

Another answer could be to simply plot it in the calculator and find the local max/min.
 
Here's a graph of your expression. By solving the quadratic which results from the differentiation, you can see the solution is among your options.
 

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If the problem wasn't multiple choice, then calculus would be needed. True to the answer might not be a miminum unless an end point, but since no domain was giver, I assume or anyone should assume all Reals.
 
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