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

Question

Elijah Abram · Accepted Answer

There is a maxima at #x=-1/3#
There is a minima at #x=1#

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 #x=-1/3#

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

There is a maxima at #x=-1/3#

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

There is a minima at #x=1#

Evelyn Agard · Answer

undefined 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} $.