# Optimization Using the First Derivative Test - Problem 1

You can find the absolute maximum of a function through a technique described earlier: by finding the critical points of a function. Remember that the critical points of a function are found by calculating the first derivative of the function and finding at what points it is equal to 0.

In order to determine if a critical point is a maximum or a minimum, you must see if the slope before and after the critical point is positive or negative. This is done by plugging in values of x that are less than or greater than the critical point into the formula for the derivative. If the slope changes from positive to negative, the point is a maximum. If the slope changes from negative to positive, the point is a minimum.

For certain functions, these critical points can also be absolute maxima or minima over the entire domain of real numbers. The function e^{x}, for example, is never less than 1.

Let’s solve an optimization problem using the first derivative test. Here I want to find the absolute max value of f(x) equals 10 x times e to the –x over 2. So the way the first derivative test works is, I’m going to take the first derivative and I’m going to look at the look for critical points. And it’ll divide the interval, this is to find over into positive and negative regions. And if I see that the derivative goes from positive to negative, I’ll have an absolute maximum.

It’s got to be positive everywhere to the left of the critical point and negative everywhere to the right. That’s an important part of the first derivative test per absolute max and min. Let’s take the derivative of this guy it’s going to require the product rule.

And that’s the first which is 10x times the derivative of the second, and that’s going to be -1/2 e to the minus x over 2. Plus the second e to the minus x over 2 times the derivative of the first which is just 10. So I can factor out let me actually multiply these two first. So I’m going to bet minus 5x e to the minus x over 2 plus 10 e to the minus x over 2.

And you can see when you look at these two terms that there is a factor of 5 e to the minus x over 2 in both terms. So I’ll pull that out. 5 e to the minus x over 2 and that leaves a –x plus 2. 2 comes from this term. So I have a completely factored derivative and it’s pretty clear that the critical point for this derivative is 2.

X equals 2 is the critical point. And there are no other critical points this piece here is never going to equal 0. And it’s also never going to be undefined. So what I’m going to do now is create the sign chart. That will help me analyze where the derivative is positive and where its negative. So remembering this derivative formula I have f'(x) equals –x plus 2 times 5e to the minus x over 2.

I’ll extend this a little bit. My one critical point x equals 2 that’s where the derivative is 0. So I want to test to the left and test to the right. If I plug in a number to the left like 0, I’ll get 2 which is positive times 5e to the it doesn’t matter, this is always going to be positive.

Positive times positive is going to be positive. And to the right say I plug in 3 I’ll have minus 2 plus 2 this will be negative times a positive it will be negative. What’s really important about the first derivative test when you are applying it for absolute max or min, as opposed to local maximum. Is that the derivative will be positive everywhere to the left over the entire domain.

We have no reason to suspect that it will ever be negative because there really is the only the one critical point. So if its positive at one point it’s going to be positive everywhere over here. And if its negative at one point over here it will be negative everywhere. So and this is really important your teacher will probably want you to justify that it’s an absolute max or min. So you really have to write out your explanation showing that you know the first derivative test. It’s because f' is greater than 0 for x less than 2.

And f' is less than 0 for x greater than 2. Because of these two things f is going to have an absolute max at x equals 2. An absolute max at x equals 2. We were actually asked for the maximum value so I need to calculate f(2).

X equals 2 is just where the maximum occurs it’s not the maximum value that’s of 2 . So going back to my original function, which was 10x so 10 times 2. E to the –x over 2 so e to the -2 over 2, I have 20e to the minus 1. Which is approximately 7.36.

That’s the maximum value for this function. It’s the biggest value it ever attains over the entire domain of real numbers.

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