The second derivative test allows you to determine the intervals on which a function is concave up or concave down by examining the sign of the second derivative around the inflexion points.
Inflexion points are points for which #f^('')(x) = 0#.
So, start by calculating the first derivative of #f(x)# - use the quotient rule
#d/dx(f(x)) = ([d/dx(5e^x)] * (5e^(x) + 6) - 5e^x * d/dx(5e^x+6))/(5e^x + 6)^2#
#f^' = (5e^x * (5e^x + 6) - 5e^x * 5e^x)/(5e^x + 6)^2#
#f^' = (color(red)(cancel(color(black)(5e^x))) + 30e^x - color(red)(cancel(color(black)(5e^x))))/(5e^x + 6)^2 = (30e^x)/(5e^x + 6)^2#
Next, calculate the second derivative by using the quotient and chain rules
#d/dx(f^'(x)) = ([d/dx(30e^x)] * (5e^x+6)^2 - 30e^x * d/dx(5e^x + 6)^2)/[(5e^x + 6)^2]^2#
#f^('') = (30e^x * (5e^x + 6)^color(red)(cancel(color(black)(2))) - 30e^x * color(red)(cancel(color(black)((5e^x + 6)))) * 5e^x)/(5e^x + 6)^color(red)(cancel(color(black)(4)))#
#f^('') = (150e^(2x) + 180e^x - 300e^(2x))/(5e^x + 6)^3 = -(30e^x(5e^x -6))/(5e^x + 6)^3#
FInd the critical poin(s) of the function by calculating
#f^('') = 0#
#-(30e^x(5e^x -6))/(5e^x + 6)^3 = 0 <=> (5e^x - 6) = 0#
This will get you
#5e^x = 6#
#e^x = 6/5 = 1.2#
Take the natual log of both sides of the equation to get
#ln(e^x) = ln(1.2) => x * lne = ln(1.2) => x = color(green)(ln(1.2))#
Now investigate the sign of the sign derivative for values smaller than #ln(1.2)# and larger than #ln(1.2)#.
SInce #e^a>0# for any real number #a#, the sign of the second derivative will depend on the numerator of the fraction, #-30e^x (5e^x - 6)#.
So, the two intervals that you're going to look at are
For values smaller than #ln(1.2)#, #(5e^x -6)# will be negative, which means that #f^('')# will be positive.
As a result, #f(x)# will be concave up on this interval.
This time, #(5e^(x) -6)>0#, so it follows that #f^('')# will be negative.
This implies that #f(x)# is concave down on this interval.
The point #(ln(1.2), f(ln(1.2)))# will be the only Inflexion point for this function.
