# What are the global and local extrema of #f(x)=8x^3-4x^2+6# ?

The local extrema are

and the global extrema are

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To find the global and local extrema of ( f(x) = 8x^3 - 4x^2 + 6 ), we first need to find its critical points by finding where its derivative equals zero. Then, we can analyze the behavior of the function around these critical points to determine the extrema.

First, we find the derivative of ( f(x) ): [ f'(x) = 24x^2 - 8x ]

Setting the derivative equal to zero to find critical points: [ 24x^2 - 8x = 0 ] [ 8x(3x - 1) = 0 ]

This gives us critical points ( x = 0 ) and ( x = \frac{1}{3} ).

Next, we need to determine the nature of these critical points by examining the sign of the derivative around them.

For ( x < 0 ), ( f'(x) ) is negative, indicating that the function is decreasing.

For ( 0 < x < \frac{1}{3} ), ( f'(x) ) is positive, indicating that the function is increasing.

For ( x > \frac{1}{3} ), ( f'(x) ) is positive, indicating that the function is increasing.

Now, we need to find the values of ( f(x) ) at these critical points and at the endpoints of the domain to determine the global extrema.

[ f(0) = 6 ] [ f\left(\frac{1}{3}\right) = \frac{622}{27} ] [ \lim_{x \to \infty} f(x) = \infty ] [ \lim_{x \to -\infty} f(x) = -\infty ]

Therefore, the global minimum occurs at ( x = 0 ) with ( f(0) = 6 ), and there is no global maximum as the function tends to positive infinity as ( x ) approaches infinity.

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