Question #2237b
1 Answer
a)
b)
c) There is no inflexion point.
Explanation:
As you have managed to find,
a) To find the interval on which
Likewise,
Therefore,
#f# is strictly increasing on#x in ]-ln7/8,oo[# , and strictly decreasing on#x in ]-oo,-ln7/8[# .
b) To find the maxima or minima (critical points), we find the
Therefore,
#x=-ln7/8# is a critical point. As we have determined in part (a) that#f# is decreasing on#x<-ln7/8# and increasing on#x> -ln7/8# , we can conclude that#x=-ln7/8# is a local minimum.
c) To find the inflexion point, at high school level or in precalculus, we find the
As
#e^x>0# for all#x inR# , there is no solution for#e^(8x)=-1/49# Therefore, there is no inflexion point for#f# .