Jason E. answered • 08/29/23
College math professor for over 10 years, tutor for over 15 years
Hi Alicia,
I am happy to answer your question. First, let us recall what the Extreme Value Theorem says, and why it is important. Then, let's apply the statement to your given problem.
The Extreme Value Theorem is a special property of continuous functions (those without jumps, holes, or infinities) that states if you restrict any continuous function to a closed interval, like [1, 10], then there is guaranteed to be a y-value (the output of the function) that is the absolute max, and also a y-value that is the absolute min. By the way, the adjective "absolute" refers to the y-value being the highest (or lowest) over all the y-values you could produce using x-values within the interval [a, b].
A few notes:
1) The absolute max and min values could change depending on the interval, meaning if you change to something different, like [-2, 1], then the answers could be different also.
2) Some functions have an overall absolute max or min that applies over its entire domain. In this case, this highest peak (or lowest valley) only counts if the x-value is within the interval [a, b] given in the problem.
3) The absolute max or min can occur at the endpoints, so those need to be checked for every problem.
So who cares?
Well, say you have a few drinks at a party. Your body will absorb the alcohol and your BAC (blood-alcohol concentration) will increase. Assuming you don't keep on drinking, at some point your body begins to metabolize that alcohol, removing it from your body. This means if we graph the BAC against time starting when you consume the alcohol, then the graph begins to increase then after some time decreases back toward zero. Then you can drive home. There is a time value for when metabolism starts, and that time occurs at the maximum of the BAC curve!
A model for such a curve has the form y = xe^(-Cx), for some constant C. This function is continuous, so if we consider over 6 hours say, then [0, 6] guarantees an absolute max we can find as the start time of your metabolism.
OK, let's get to your problem!
First thing I notice immediately is that 4/4x can be reduced to 1/x, so why not just do that?
So we have y = x^2 + 1/x over the closed interval [1, 10].
First, the endpoints:
x = 1 implies y = 1^2 + 1/1 = 2
x = 10 implies y = 10^2 + 1/10 = 100.1
Second, the calculus to find any max or min values in between x = 1 and x = 10. To do so, we find the derivative y' and solve y' = 0. Both derivatives of x^2 and 1/x can found by the Power Rule, but 1/x needs to first be rewritten as x^(-1). That negative exponent just brings the x from downstairs to upstairs. The Power Rule only applies to x's that are upstairs. So
y ' = 2x -x^(-2) = 2x - 1/x^2
After taking the derivative, it is good practice to rewrite all terms back with positive exponents. Because next come the algebra to solve y' = 0:
y' = 0 = 2x - 1/x^2
There are several ways to solve for x. I encourage you to try it yourself. Here is one way to do it:
add 1/x^2 to get
2x = 1/x^2
multiply by x^2 to get
2x^3 = 1
and so x = cube root(1/2) = 0.794! (recall odd roots yield only one answer)
This imples y = (0.794)^2 + 1/0.794 = 1.890.
Finally, let's put it all together: possibilities for absolute max and min are
x = 1 implies y = 1^2 + 1/1 = 2
x = 10 implies y = 10^2 + 1/10 = 100.1
x = 0.794 implies y = 1.890
From this, we must have the absolute min occurs at x = 0.794 with min value y = 1.890 and the absolute max occurs at the right endpoint x = 10 with max value y = 100.1. The other endpoint x = 2 is nothing.
I hope that helps! Please reach out to me if you would like further help in your class or if you have any questions about my solution.
