For the function, find all points of inflection or determine that no such points exist.

Answer

We don't need to use the quotient rule. First, rewrite the function:

*f* (*x*) = (1 + 5*e*^{-x})^{-1}

Now use the chain rule to take the derivative:

To take the second derivative we need to use the product rule and the chain rule:

This is the second derivative. But it's ugly. Can we make it more clear? Since -2 = -3 + 1, we can replace

(1 + 5*e*^{-x})^{-2}

with

(1 + 5*e*^{-x})^{-3}(1 + 5*e*^{-x}).

Believe us? Good. This is what we have now:

Phew! This looks better, since it's all broken down into factors.

*f*" can never be undefined. Is it ever zero? Go through one factor at a time.

*f*"(*x*) = -*5**e*^{-x} (1 + 5*e*^{-x})^{-3}[-5*e*^{-x} + 1]

The factor 5*e*^{-x} is never zero.

*f*"(*x*) = -5*e*^{-x}*(1 + 5**e*^{-x})^{-3}[-5*e*^{-x} + 1]

Since 5*e*^{-x} is always positive, so is (1 + 5*e*^{-x}), and so is (1 + 5*e*^{-x})^{-3}. That part is never zero.

*f*"(*x*) = -5*e*^{-x} (1 + 5*e*^{-x})^{-3}*[-5**e*^{-x} + 1]

This part can equal zero if

0 = -5*e*^{-x} + 1.

Solve:

Since *ln e*= 1, therefore

Remember, this is still only a possible IP. Plug in numbers to the left and right of our possible IP, and see what happens to the sign of *f*". We already know the only factor of *f*" that isn't always positive is the last one:

Therefore if the sign of [-5*e*^{-x} + 1] changes from one side to the other of our possible IP, the sign of *f*" changes also, and so we really do have an inflection point.

For a point to the left we'll use *x* = 1, and for a point to the right use *x* = 2. Pull out the calculator:

The sign of [-5*e*^{-x} + 1] changes from one side to the other of our possible IP, so we really do have an IP at