How do use the first derivative test to determine the local extrema #f(x) = (x+3)(x-4)^2#?

Question

Elijah Abram · Accepted Answer

See the explanation.

#f(x) = (x+3)(x-4)^2# #f'(x) = [1] (x-4)^2 + (x+3)[2(x-4)(1)]# # = (x-4)[(x-4)+(x+3)2]# # = (x-4)(3x+2)# The critical numbers for #f# are #4 " and "-2/3# We look at the sign of #f'# on each interval to determine whether #f# is increasing or decreasing on the interval #{: (bb "Interval", bb"Sign of "f',bb" Incr/Decr"), ((-oo,-2/3)," " +" ", " "" Incr"), ((-3/2,4), " " -, " " " Decr"), ((4,oo), " " +, " "" Incr") :}# #f# has a local maximum of #1372/27# at #-2/3# and a local minimum of #0# at #4#.

Christopher Agresta · Answer

undefined To use the first derivative test to determine the local extrema of $ f(x) = (x+3)(x-4)^2 $:

1. Find the first derivative of the function $ f(x) $ using the product rule.
2. Set the first derivative equal to zero and solve for $ x $ to find critical points.
3. Test the intervals between the critical points by evaluating the sign of the first derivative within each interval.
4. If the sign changes from positive to negative, there is a local maximum; if it changes from negative to positive, there is a local minimum; if there is no sign change, there is neither a local maximum nor a local minimum.