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

Question

Isaiah Allen · Accepted Answer

That function has no global extrema. It has local maximum of #8# (at (#x=-1#) and local minimum of #4# (at #x=1#)

#lim_(xrarroo)f(x) = oo#, so there is no global maximum. #lim_(xrarr-oo)f(x) = -oo#, so there is no global minimum. #f'(x) = 3x^2-3# which is never undefined and is #0# at #x=-1# and at #x=1#. The domain of #f# is #RR#. Therefore, the only critical numbers are #-1# and #1#. The sign of #f'# changes from + to - as we pass #x=-1#, so #f(-1) = 8# is a local maximum. The sign of #f'# changes from - to + as we pass #x=1#, so #f(1) = 4# is a local minimum.

Cameron Alexander · Answer

undefined To find the global and local extrema of $ f(x) = x^3 - 3x + 6 $, we first find the critical points by setting the derivative equal to zero and solving for $ x $.

Taking the derivative of $ f(x) $, we get $ f'(x) = 3x^2 - 3 $. Setting $ f'(x) $ equal to zero and solving for $ x $, we find:

$$ 3x^2 - 3 = 0 $$
$$ x^2 - 1 = 0 $$
$$ (x-1)(x+1) = 0 $$

So, $ x = 1 $ and $ x = -1 $ are the critical points.

To determine whether these critical points are local minima, local maxima, or neither, we can use the second derivative test.

Taking the second derivative of $ f(x) $, we get $ f''(x) = 6x $.

At $ x = 1 $, $ f''(1) = 6 $ which is positive, indicating a local minimum.

At $ x = -1 $, $ f''(-1) = -6 $ which is negative, indicating a local maximum.

Since there are no other critical points and the function is a cubic polynomial, it extends indefinitely in both directions. Therefore, there is no global maximum or minimum.