SOLUTION: Determine the location of each extremum of the function f(x)= x^3-1/2x^2-6x+4 f(x)= 3x^2-x-6 = x^2-x/3-2=0 that's all i got

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Question 998301: Determine the location of each extremum of the function
f(x)= x^3-1/2x^2-6x+4
f(x)= 3x^2-x-6
= x^2-x/3-2=0 that's all i got
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
Determine the location of each extremum of the function
f(x)= x^3-1/2x^2-6x+4
f(x)= 3x^2-x-6 ( = f'(x), not f(x) )
= x^2-x/3-2=0 that's all i got
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Set f' = 0
3x^2 - x - 6 = 0
The extrema are at the zeroes of f'(x)
Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 3x%5E2%2B-1x%2B-6+=+0) has the following solutons:

x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%28-1%29%5E2-4%2A3%2A-6=73.

Discriminant d=73 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--1%2B-sqrt%28+73+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%28-1%29%2Bsqrt%28+73+%29%29%2F2%5C3+=+1.59066729088626
x%5B2%5D+=+%28-%28-1%29-sqrt%28+73+%29%29%2F2%5C3+=+-1.25733395755292

Quadratic expression 3x%5E2%2B-1x%2B-6 can be factored:
3x%5E2%2B-1x%2B-6+=+%28x-1.59066729088626%29%2A%28x--1.25733395755292%29
Again, the answer is: 1.59066729088626, -1.25733395755292. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+3%2Ax%5E2%2B-1%2Ax%2B-6+%29

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Those are the x values for the max (negative value) and min (positive value).
Finding the y values is tedious, but just algebra.