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

Local extrema:

Global extrema:

Local extrema, also called maxima & minima, or sometimes critical points, are just what they sound like: when the function's reached a brief maximum or a brief minimum. They're called local because when you're looking for critical points, you usually only care about what the maximum means in the immediate neighborhood of the point.

Finding local critical points are pretty simple. Find when the function is unchanging, and the function is unchanging when - you guessed it - the derivative equals zero.

We're concerned when this expression equals zero:

So we've determined that there are two local extrema, as well as their locations. Classifying whether each point is a maximum or minimum is a different story and I won't go into that here, but I can direct you here if that's something you'd like to read up on.

With global extrema, there's more you need to take into account than just the derivative. You would have to determine if there are any critical points on this interval, because if so, one might (but not necessarily) be the global extrema as well. With these types of situations, having a calculator plot is the most helpful, but a little analysis reveals the critical points. (I can direct you to this page for more info and a few examples)

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To find the global and local extrema of ( f(x) = x^3 - x^2 - x + 1 ), we first find the critical points by setting the derivative equal to zero and then classify them using the second derivative test.

[ f'(x) = 3x^2 - 2x - 1 ]

Setting ( f'(x) = 0 ), we solve for critical points:

[ 3x^2 - 2x - 1 = 0 ]

Using the quadratic formula:

[ x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(3)(-1)}}{2(3)} ]

[ x = \frac{2 \pm \sqrt{4 + 12}}{6} ]

[ x = \frac{2 \pm \sqrt{16}}{6} ]

[ x = \frac{2 \pm 4}{6} ]

[ x_1 = \frac{1}{3} ] [ x_2 = -1 ]

To classify the critical points, we find the second derivative ( f''(x) ):

[ f''(x) = 6x - 2 ]

Evaluate ( f''(x) ) at the critical points:

[ f''\left(\frac{1}{3}\right) = 6\left(\frac{1}{3}\right) - 2 = 0 ] [ f''(-1) = 6(-1) - 2 = -8 ]

At ( x = \frac{1}{3} ), the second derivative test is inconclusive. At ( x = -1 ), since ( f''(-1) < 0 ), it's a local maximum.

To find the global extrema, we evaluate ( f(x) ) at the critical points and endpoints of the interval:

[ f\left(\frac{1}{3}\right) = \left(\frac{1}{3}\right)^3 - \left(\frac{1}{3}\right)^2 - \frac{1}{3} + 1 = \frac{7}{27} ] [ f(-1) = (-1)^3 - (-1)^2 - (-1) + 1 = 4 ] [ f(0) = 0^3 - 0^2 - 0 + 1 = 1 ]

Since ( f(-1) = 4 ) is the maximum among these values, it is the global maximum. There is no global minimum because the function is unbounded below.

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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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