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.