Question

Question #2237b

Answer 1

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`.