What are the absolute extrema of #f(x) =x^4 − 8x^2 − 12 in[-3,-1]#?

Question

Evelyn Agard · Accepted Answer

#-3# (occurring at #x=-3#) and #-28# (occurring at #x=-2#)

Absolute extrema of a closed interval occur at the endpoints of the interval or at #f'(x)=0#. That means we'll have to set the derivative equal to #0# and see what #x#-values that gets us, and we'll have to use #x=-3# and #x=-1# (because these are the endpoints). So, starting with taking the derivative: #f(x)=x^4-8x^2-12# #f'(x)=4x^3-16x# Setting it equal to #0# and solving: #0=4x^3-16x# #0=x^3-4x# #0=x(x^2-4)# #x=0# and #x^2-4=0# Thus the solutions are #0,2,# and #-2#. We immediately get rid of #0# and #2# because they are not on the interval #[-3,-1]#, leaving only #x=-3,-2,# and #-1# as the possible places where extrema can occur. Finally, we evaluate these one by one to see what the absolute min and max are: #f(-3)=-3# #f(-2)=-28# #f(-1)=-19# Therefore #-3# is the absolute maximum and #-28# is the absolute minimum on the interval #[-3,-1]#.

Paisley Johnson · Answer

undefined To find the absolute extrema of $ f(x) = x^4 - 8x^2 - 12 $ in the interval $[-3, -1]$, we need to evaluate the function at the critical points and endpoints within the given interval.

1. Find critical points by setting the derivative equal to zero and solving for $ x $.
$$ f'(x) = 4x^3 - 16x = 0 $$
$$ 4x(x^2 - 4) = 0 $$
$$ x = 0, \pm 2 $$

2. Evaluate $ f(x) $ at these critical points and endpoints.
$$ f(-3) = (-3)^4 - 8(-3)^2 - 12 = 81 - 72 - 12 = -3 $$
$$ f(-1) = (-1)^4 - 8(-1)^2 - 12 = 1 - 8 - 12 = -19 $$
$$ f(0) = -12 $$
$$ f(2) = 2^4 - 8(2)^2 - 12 = 16 - 32 - 12 = -28 $$

3. Compare these values to determine the absolute extrema.
The minimum value is $ -28 $ and the maximum value is $ -3 $.

Therefore, the absolute minimum of $ f(x) $ in $[-3, -1]$ is $ -28 $ at $ x = 2 $, and the absolute maximum is $ -3 $ at $ x = -3 $.