Science:Math Exam Resources/Courses/MATH100/December 2019/Question 6

MATH100 December 2019

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Question 6
Find, with proof, the $(x,y)$ coordinates of the global maximum and global minimum of the function $f(x)=x\cdot {\sqrt {6-x}}$ on the closed interval $[-2,5]$.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1
What are some qualities which a function $f(x)$ must have at a point $x_{0}$ in its domain in order for $f(x_{0})$ to be a local maximum or minimum? Note: this will involve the the derivative, but do not forget potential other points.

Hint 2
Notice that the domain of our function is a closed interval. So, besides critical points, be sure to also check the endpoints of the domain of our function.

Hint 3
Since this is asking for a proof, try to give explanations for your calculations. This helps demonstrate to the reader the reasoning behind your proof.

Checking a solution serves two purposes: helping you if, after having used all the hints, you still are stuck on the problem; or if you have solved the problem and would like to check your work. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. We are asked to find the global maximum and global minimum of the function $f(x)=x\cdot {\sqrt {6-x}}$ on the closed interval $[-2,5]$. Because the domain of our function is a closed interval, we need to check the left-hand and right-hand endpoints, as well as critical points. These are the points $x_{0}$ in the domain of $f$ where $f^{'}(x_{0})=0$, or where the derivative of $f$ fails to exist. Let us first identify critical points. First, since $-2\leq x\leq 5$, the function $x\cdot {\sqrt {6-x}}$ is well-defined (i.e. makes sense) and differentiable. Moreover, utilizing the product and power rules, we see that: $f^{'}(x)={\sqrt {6-x}}-{\frac {x}{2{\sqrt {6-x}}}}$. We now assume that $x_{0}$ is a real number, with $-2\leq x_{0}\leq 5$, and $f^{'}(x_{0})=0$. Then, ${\sqrt {6-x_{0}}}-{\frac {x_{0}}{2{\sqrt {6-x_{0}}}}}=0\Rightarrow {\sqrt {6-x_{0}}}={\frac {x_{0}}{2{\sqrt {6-x_{0}}}}}\Rightarrow 2(6-x_{0})=x_{0}\Rightarrow x_{0}=4$. So, we have $x_{0}$ is the only critical point of the function $f$ in the interval $[-2,5]$. Now, let us test our three points $-2,4,5$ by plugging them into our function $f(x)=x\cdot {\sqrt {6-x}}$. We have: $f(-2)=-2{\sqrt {6-(-2)}}=-4{\sqrt {2}},\quad f(4)=4{\sqrt {6-4}}=4{\sqrt {2}},\quad f(5)=5{\sqrt {6-5}}=5.$ From the above, on the interval $[-2,5]$, $f(x)$ has global minimum $(-2,-4{\sqrt {2}})$, and global maximum at the point $(4,4{\sqrt {2}})$. This is because ${\sqrt {2}}\approx 1.5\Rightarrow 4\times {\sqrt {2}}>5$.