How do you use the second derivative test to find where the function #f(x) = (5e^x)/(5e^x + 6)# is concave up, concave down, and inflection points?

Question

Christopher Agresta · Accepted Answer

#f(x)# will be concave down on #(ln(1.2), +oo)# and concave up on #(-oo, ln(1.2))#.

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.

Ezra Adams · Answer

undefined To use the second derivative test for the function $ f(x) = \frac{5e^x}{5e^x + 6} $ to find where it is concave up, concave down, and its inflection points, follow these steps:

1. Find the first derivative of $ f(x) $ with respect to $ x $ to get $ f'(x) $.
2. Find the second derivative of $ f(x) $ by differentiating $ f'(x) $ with respect to $ x $ to get $ f''(x) $.
3. Set $ f''(x) = 0 $ and solve for $ x $ to find any points where the concavity may change, these are potential inflection points.
4. Determine the concavity of the function by analyzing the sign of $ f''(x) $ in intervals between critical points found in step 3.
5. Use the second derivative test:
   - If $ f''(x) > 0 $ for a given interval, the function is concave up on that interval.
   - If $ f''(x) < 0 $ for a given interval, the function is concave down on that interval.
6. Confirm potential inflection points by checking the concavity on either side of them.

These steps will help you identify where the function $ f(x) $ is concave up, concave down, and its inflection points.