Question #2237b

1 Answer
Jun 13, 2017

a) #f# is strictly increasing on #x in ]-ln7/8,oo[#, and strictly decreasing on #x in ]-oo,-ln7/8[#.
b) #x=-ln7/8# is a local minimum.
c) There is no inflexion point.

Explanation:

#f(x)=e^(7x)+e^(-x)#
As you have managed to find, #f'(x)=7e^(7x)-e^(-x)#

a) To find the interval on which #f# is increasing or decreasing, we find the interval on which #f'(x) > or < 0#
#f'(x)>0#
#7e^(7x)>e^(-x)#
#7e^(8x)>1#
#ln(e^(8x))>ln(1/7)#
#8x> -ln7#
#x> -ln7/8#
Likewise,
#f'(x)<0#
#7e^(7x)< e^(-x)#
#7e^(8x)< 1#
#ln(e^(8x))< ln(1/7)#
#8x< -ln7#
#x< -ln7/8#

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 #x# such that #f'=0# at those points.
#f'(x)=0#
#7e^(7x)=e^(-x)#
#7e^(8x)=1#
#ln(e^(8x))=ln(1/7)#
#8x=-ln7#
#x=-ln7/8#

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 #x# such that #f''=0# at those points.
#f'(x)=7e^(7x)-e^(-x)#
#f''(x)=49e^(7x)+e^(-x)#
#f''(x)=0#
#49e^(7x)+e^(-x)=0#
#e^(8x)=-1/49#

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#.