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.
