Question

Find absolute maximum and minimum of f(x)=ax+(b/x^2), x>0 when a,b>0

Answer 1

\[f(x)=ax+\frac{b}{x^2}\\
\begin{align*}f'(x)=a-\frac{2b}{x^3}&=0\\
a&=\frac{2b}{x^3}\\
x^3&=\frac{2b}{a}\\
x&=\sqrt[3]\frac{2b}{a}
\end{align*}\]
Apply the first derivative test to determine how \(f(x)\) behaves before/after this value of \(x\).

Answer 2

Wait, so how do I do that? Would I plug in what x is into the original equation?

Answer 3

Doing so would give you the extreme value that the function attains. You must first verify that this value indeed gives an extremum.

Say you have some function \(f(x)\), and \(f'(x)=0\) when \(x=c\). This \(c\) is called a critical point. Now, depending on the domain of \(f(x)\), you would check the sign on the derivative of \(f(x)\) before and after \(x=c\).

Let's try an example. Suppose \(f(x)=x^2\), a parabola with a minimum value at its vertex, which occurs when \(x=0\). You have that
\[f'(x)=2x=0~~\Rightarrow~~x=0\]
Now, check the behavior of the function to the left and right of 0 by picking some convenient value less than or greater than (respectively) 0:
\[\begin{align*}\text{left: }&f'(-1)=2(-1)=-2<0~~&\Rightarrow&~~f(x)\text{ is decreasing on }(-\infty,0)\\
\text{right: }&f'(1)=2(1)=2>0~~&\Rightarrow&~~f(x)\text{ is increasing on }(0,\infty)\end{align*}\]
The decrease-increase pattern indicates that a minimum indeed occurs.

Answer 4

I get that, the thing that is confusing is that it contains constants instead of numbers, so I don't know which value to test for each to determine if it's increasing or decreasing.

Answer 5

The tricky part about your question is that you're given unknown constants \(a,b\). Let's modify the procedure a bit. You're given that \(x>0\), which is your domain. This means that, since \(a\) and \(b\) are also positive, \(\sqrt[3]{\dfrac{2b}{a}}\) is strictly non-zero. What this says is that you have two intervals to test:
\[\left(0,\sqrt[3]\frac{2b}{a}\right)~~\text{and}~~\left(\sqrt[3]\frac{2b}{a},\infty\right)\]
Picking a test value for each interval is somewhat tricky, too, particularly for the second interval. You can just pick a number one less than \(\sqrt[3]{\dfrac{2b}{a}}\); this is because if \(\sqrt[3]{\dfrac{2b}{a}}=1\) (i.e. \(a=2,b=1\)), then one less would be 0, which is not in the domain.

Answer 6

I see I've made a mention of your concerns :)

What you could do is pick some arbitrarily small number; let's call it \(k\). The "arbitrarily small" part means that we can pick any number \(k\) that makes \(\sqrt[3]{\dfrac{2b}{a}}-k>0\).

Answer 7

Okay, so how would you determine the absolute maximum and minimum from that?

Answer 8

\(\color{blue}{\text{Originally Posted by}}\) @SithsAndGiggles
So I used the derivative from my example... Problem solved.
\[f'\left(\sqrt[3]\frac{2b}{a}\right)=a-\frac{2b}{\sqrt[3]\frac{2b}{a}-k}\]
Clearly, the second term is positive. You can show that \(a\) is \(\bf smaller\) than the second term, which then means that \(f'\left(\sqrt[3]{\dfrac{2b}{a}}\right)\) is negative. So, \(f(x)\) is decreasing over this interval.
\(\color{blue}{\text{End of Quote}}\)

Answer 9

\(\color{blue}{\text{Originally Posted by}}\) @SithsAndGiggles
Well, you have to verify that \(f'\) is negative, then you have to show that \(f'\) is positive on the other interval. If that's true, then a maximum occurs at the critical value you found. The extremum value would be the function evaluated at the critical point.
\(\color{blue}{\text{End of Quote}}\)

Answer 10

Another correction, it should be\[a-\frac{b}{\left(\sqrt[3]\frac{2b}{a}-k\right)^3}\]

Answer 11

Oh alright! That makes so much more sense! Thank you so much!!

Answer 12

yw