Video Transcript
What is the minimum value of the function 𝑓 of 𝑥 equals negative two 𝑥 multiplied by 𝑒 to the power of negative 𝑥?
Recall that the minimum value of a function is the smallest 𝑦-value that the function attains. Similarly, the maximum value of a function is the largest 𝑦-value that the function attains. In order to find the minimum or maximum values of a function 𝑓, we generally carry out the following procedure. Firstly, we find the first derivative of 𝑓, which is denoted by 𝑓 prime of 𝑥 or 𝑓 dash of 𝑥. We then equate the first derivative of 𝑥 to zero and solve for 𝑥. Next, we find the second derivative of 𝑓, which is denoted by 𝑓 double prime of 𝑥 or 𝑓 double dash of 𝑥. Ultimately, we substitute the solutions to the equation 𝑓 prime of 𝑥 equals zero that we find in step two into the second derivative of 𝑓. If the result is a positive number, then the solution corresponds to a minimum value. If the result is a negative number, then the solution corresponds to a maximum value.
Let’s apply this procedure to the function given to us in the question in order to find its minimum value. First, let’s find the first derivative of the function 𝑓 given to us in the question. Note that the function 𝑓 is a product of two functions, negative two 𝑥 and 𝑒 to the power of negative 𝑥. So in order to differentiate it, we will use the product rule. Using the product rule, we obtain that the derivative of 𝑓 is the derivative of negative two 𝑥, which is negative two, multiplied by 𝑒 to the power of negative 𝑥 plus negative two 𝑥 multiplied by the derivative of 𝑒 to the power of negative 𝑥, which is negative 𝑒 to the power of negative 𝑥, using the chain rule. This simplifies to negative two multiplied by 𝑒 to the power of negative 𝑥 plus two 𝑥 multiplied by 𝑒 to the power of negative 𝑥. Since the term two 𝑒 to the power of negative 𝑥 is common in both summands, we can factor it out.
Now, let’s solve the equation 𝑓 prime of 𝑥 equals zero. Doing so, we obtain that either 𝑥 minus one equals zero or 𝑒 to the power of negative 𝑥 equals zero. Now, recall that 𝑒 to the power of negative 𝑥 equals zero has no solutions. You can remember this by noting that the graph of the function 𝑦 equals 𝑒 to the power of negative 𝑥 does not cross the 𝑥-axis. So we obtain that 𝑥 equals one is the only solution to the equation 𝑓 prime of 𝑥 equals zero.
Now, let’s check that the value 𝑥 equals one actually does correspond to a minimum value as suggested by the question. Let’s compute the second derivative of 𝑓. To do this, we will differentiate the first derivative of 𝑓 using the product rule. Doing so, we obtain the derivative of two 𝑒 to the power of negative 𝑥, which is negative two 𝑒 to the power of negative 𝑥, multiplied by 𝑥 minus one plus two 𝑒 to the power of negative 𝑥 multiplied by the derivative of 𝑥 minus one, which is just one.
Now, let’s substitute 𝑥 equals one into the second derivative of 𝑓. Doing so, we obtain zero plus two 𝑒 to the power of negative one, which is just two 𝑒 to the power of negative one. Since two 𝑒 to the power of negative one is a positive number, the criteria from step four tells us that the function 𝑓 evaluated at 𝑥 equals one is a minimum value.
Evaluating the function 𝑓 at 𝑥 equals one, we obtain the value negative two over 𝑒, which is its minimum value.
