Question

Determine maximum and minimum of the function
\(\large{f(x)} = \large{\frac{2-x}{x^{2}+12}} \)
on the interval \(I=[2,12]\).

Answer 1

First, you need to take the derivative of the function. You could do this using the quotient rule.

Answer 2

it was something like that right ?
\[|\frac{a_{n+1}}{a_{n}}|\]

Answer 3

Using this, you get that the derivative is \[\frac{(x^2+12)(-1)-(2-x)(2x)}{(x^2+12)^2}=\frac{x^2-4x-12}{(x^2+12)^2}\]

Answer 4

Now, we want to find the critical points since that is where we can have maximum and minimum values. We find these by setting our derivative equal to 0. \[\frac{x^2-4x-12}{(x^2+12)^2}=0\]Factor the numerator,\[\frac{(x-6)(x+2)}{(x^2+12)^2}=0\]So this means that\[(x-6)(x+2)=0\]

Answer 5

Hence, \(x=-2,\,\,6\). However, our interval is from [2, 12] so we can throw out \(x=-2\), leaving us only with \(x=6\).

Answer 6

Now, we plug this back into the original function, to get\[\frac{2-6}{6^{2}+12}=\frac{-4}{48}=-\frac{1}{12}\]But we're not done yet. This may be the min or the max, but we need to make sure, and figure out which one it is.

Answer 7

Fortunately, we only need to test two more values. In particular, these values are the ones at the edge of the interval. So we check what \(f(2)\) and \(f(12)\) are. \[f(2)=\frac{2-2}{2^{2}+12}=\frac{0}{16}=0\]\[f(12)=\frac{2-12}{12^{2}+12}=\frac{-10}{156}=-\frac{5}{78}\]

Answer 8

Since \(-\frac{1}{12}<-\frac{5}{78}<0\) we have that the minimum is at \((6, -\frac{1}{12})\) and the maximum is at \((2, 0)\).

Answer 9

cool solution very clear, even i start to understand it, thank you very much George

Answer 10

You're welcome.