# What are the global and local extrema of #f(x)=x^3-3x+6# ?

That function has no global extrema. It has local maximum of

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To find the global and local extrema of ( f(x) = x^3 - 3x + 6 ), we first find the critical points by setting the derivative equal to zero and solving for ( x ).

Taking the derivative of ( f(x) ), we get ( f'(x) = 3x^2 - 3 ). Setting ( f'(x) ) equal to zero and solving for ( x ), we find:

[ 3x^2 - 3 = 0 ] [ x^2 - 1 = 0 ] [ (x-1)(x+1) = 0 ]

So, ( x = 1 ) and ( x = -1 ) are the critical points.

To determine whether these critical points are local minima, local maxima, or neither, we can use the second derivative test.

Taking the second derivative of ( f(x) ), we get ( f''(x) = 6x ).

At ( x = 1 ), ( f''(1) = 6 ) which is positive, indicating a local minimum.

At ( x = -1 ), ( f''(-1) = -6 ) which is negative, indicating a local maximum.

Since there are no other critical points and the function is a cubic polynomial, it extends indefinitely in both directions. Therefore, there is no global maximum or minimum.

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