# What are the local extrema of #f(x)= x^3 - 3x^2 - x + 1#?

A polynomial is continuous and has a continuous derivative, so the extrema can be found by equating the derivative function to zero and solving the resulting equation.

The derivative function is

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To find the local extrema of the function ( f(x) = x^3 - 3x^2 - x + 1 ), we first find its critical points by taking the derivative of the function and setting it equal to zero. Then, we determine whether these critical points correspond to local maxima or minima by analyzing the sign changes of the derivative around them.

Taking the derivative of ( f(x) ), we get: [ f'(x) = 3x^2 - 6x - 1 ]

Setting ( f'(x) ) equal to zero and solving for ( x ), we find the critical points: [ 3x^2 - 6x - 1 = 0 ] [ x = \frac{6 \pm \sqrt{36 + 12}}{6} ] [ x = \frac{6 \pm \sqrt{48}}{6} ] [ x = \frac{6 \pm 4\sqrt{3}}{6} ] [ x = 1 \pm \frac{2\sqrt{3}}{3} ]

Now, we evaluate the second derivative of ( f(x) ) at these critical points to determine the concavity: [ f''(x) = 6x - 6 ]

Plugging in the critical points: [ f''\left(1 + \frac{2\sqrt{3}}{3}\right) = 6\left(1 + \frac{2\sqrt{3}}{3}\right) - 6 = 6\sqrt{3} - 6 > 0 ] [ f''\left(1 - \frac{2\sqrt{3}}{3}\right) = 6\left(1 - \frac{2\sqrt{3}}{3}\right) - 6 = -6\sqrt{3} - 6 < 0 ]

Since ( f''\left(1 + \frac{2\sqrt{3}}{3}\right) > 0 ), the critical point ( x = 1 + \frac{2\sqrt{3}}{3} ) corresponds to a local minimum. And since ( f''\left(1 - \frac{2\sqrt{3}}{3}\right) < 0 ), the critical point ( x = 1 - \frac{2\sqrt{3}}{3} ) corresponds to a local maximum.

Therefore, the local extrema of ( f(x) = x^3 - 3x^2 - x + 1 ) are:

- Local maximum at ( x = 1 - \frac{2\sqrt{3}}{3} )
- Local minimum at ( x = 1 + \frac{2\sqrt{3}}{3} )

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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