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

There is a maxima at

There is a minima at

Given -

#y=x^3-x^2-x#

#dy/dx=3x^2-2x-1#

#(d^2y)/(dx^2)=6x-2#

Set the first derivative equal to zero

#dy/dx=0 =>3x^2-2x-1=0#

#3x^2-3x+x-1=0#

#3x(x-1)+1(x-1)=0#

#(3x+1)(x-1)=0#

#3x=-1#

#x=-1/3#

#x-1=0#

#x=1#

At

#(d^2y)/(dx^2)=6(-1/3)-2=-2-2=-4<0#

There is a maxima at

#(d^2y)/(dx^2)=6(1)-2=6-2=4>0#

There is a minima at

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To find the global and local extrema of ( f(x) = x^3 - x^2 - x ), we first take the derivative of the function, set it equal to zero to find critical points, and then analyze the behavior of the function around these points.

Taking the derivative: [ f'(x) = 3x^2 - 2x - 1 ]

Setting ( f'(x) ) equal to zero and solving for ( x ) gives us critical points: [ 3x^2 - 2x - 1 = 0 ]

Using the quadratic formula, we find: [ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} ]

Where ( a = 3 ), ( b = -2 ), and ( c = -1 ). [ 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} ]

So, the critical points are: [ x_1 = \frac{1}{3} ] [ x_2 = -1 ]

Now, we need to test the behavior of the function around these critical points to determine if they correspond to local extrema.

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

Since the second derivative test is inconclusive, we'll resort to the first derivative test. Evaluating ( f'(x) ) on either side of ( x = \frac{1}{3} ): [ f'\left(0\right) = -1 ] [ f'\left(\frac{1}{2}\right) = \frac{1}{4} ]

Since the derivative changes sign from negative to positive at ( x = \frac{1}{3} ), ( f(x) ) has a local minimum at ( x = \frac{1}{3} ).

For ( x = -1 ): [ f''(-1) = 6(-1) - 2 ] [ f''(-1) = -6 - 2 ] [ f''(-1) = -8 ]

Since the second derivative is negative, ( f(x) ) has a local maximum at ( x = -1 ).

To find the global extrema, we evaluate the function at the critical points and endpoints of the domain, which is all real numbers: [ f\left(\frac{1}{3}\right) = \frac{1}{27} - \frac{1}{9} - \frac{1}{3} = -\frac{7}{27} ] [ f(-1) = (-1)^3 - (-1)^2 - (-1) = -1 + 1 + 1 = 1 ]

Since ( f(-1) = 1 ) is greater than ( f\left(\frac{1}{3}\right) = -\frac{7}{27} ), the global maximum is ( f(-1) = 1 ) and the global minimum is ( f\left(\frac{1}{3}\right) = -\frac{7}{27} ).

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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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