# F(x)=2x + 32/x find the absolute maximum and minimum if they exist

• AFNequation
In summary, the absolute maximum and absolute minimum for the given function is when x > 0 and when x = 0.

#### AFNequation

find the absolute maximum and absolute minimum (if any) for the given function :
f(x)= 2x + 32/x ; when x > 0

I tried to solve it but I'm not sure if its a maxima or minima, here are my try :
f(x)= 2x + 32/x
f'(x)= 2-32/x^2
2-32/x^2=0
2u^2=32
u^2=16
u=4 or - 4
-4 is not included cause x > 0

f(4)= 2(4) + 32/4
f(4)= 8 + 8
f(4)= 16
(4, 16) but I'm not sure if it an absolute maximum or minimum?

are my steps correct ?

Find the second derivative at x = 4 to determine if it's a minimum or a maximum. If f''(4) > 0, then it's a local minimum, and if f''(4) < 0 it's a local maximum. Since the only x satisfying x > 0 and f'(x) = 0 is at x = 4, you can tell that it must be an absolute extremum.

thank you
but isn't the second derivative test only to determine the max or min relative ? not absolute extrema?

Yes, but if you find only one local maximum and no other places where f'(x) = 0, it must be an absolute maximum. Try drawing a graph. Draw a local maximum, and try to continue the curve and make it attain another higher maximum. Can you get it to do that without creating a local minimum somewhere?

i got another quistion
do i have to find f(-4) and f''(-4) ? or is it just f(4) and f''(-4) ?
cause it is > o

The problem as given asked for the maximum and minimum of your function for x > 0, so no, you don't need to check x = -4.

AFNequation said:
find the absolute maximum and absolute minimum (if any) for the given function :
f(x)= 2x + 32/x ; when x > 0

I tried to solve it but I'm not sure if its a maxima or minima, here are my try :
f(x)= 2x + 32/x
f'(x)= 2-32/x^2
2-32/x^2=0
2u^2=32
u^2=16
u=4 or - 4
-4 is not included cause x > 0

f(4)= 2(4) + 32/4
f(4)= 8 + 8
f(4)= 16
(4, 16) but I'm not sure if it an absolute maximum or minimum?

are my steps correct ?

Let's take a step back and recall something important.

A function is guaranteed to have an absolute maximum and an absolute minimum on a closed and bounded (aka compact) interval [a,b] if it is continuous everywhere in that interval.

Now, your function is continuous for all x > 0, but you are looking for a maximum and minimum on the interval (0, infinity), and this interval is NOT compact. Therefore the function is NOT guaranteed to have either a maximum or a minimum on that interval!

In fact, it certainly has no absolute maximum: if you plug in increasingly smaller numbers for x, then 2x gets closer and closer to 0, but 32/x gets larger and larger without bound! Similarly, if you pick x larger and larger, 32/x gets closer and closer to zero, but 2x grows without bound. Thus I could make f(x) equal a trillion or a googol or any big number I like if I choose a small enough x or a large enough x.

So at best you can hope to find an absolute minimum.

To verify that an absolute maximum does exist, let's pick x1 and x2 so that

$$f(x1) \geq 1000000000000$$ for $$0 \leq x < x1$$ or $$x > x2$$

Then $$[x1,x2]$$ is a compact interval, and f is continuous at every point in that interval, so it has an absolute minimum (and absolute maximum) when restricted to that interval. As long as the absolute minimum on $$[x1,x2]$$ is less than 1000000000000, then it is also the absolute minimum on all of $$(0,\infty)$$.

Since f is also differentiable on all of $$[x1,x2]$$, you can now proceed with the usual first and second derivative tests (along with checking the endpoints) to find the absolute max and min on $$[x1,x2]$$.

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