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 onx in ]-ln7/8,oo[ , and strictly decreasing onx 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) thatf is decreasing onx<-ln7/8 and increasing onx> -ln7/8 , we can conclude thatx=-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 allx inR , there is no solution fore^(8x)=-1/49 Therefore, there is no inflexion point forf .