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

[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 ?

 

[dx]

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.

 

[AFNequation]

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

 

[dx]

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?

 

[AFNequation]

thanks for your help :)

 

[AFNequation]

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

 

[Mark44]

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.

 

[jbunniii]

find the absolute maximum and absolute minimum (if any) for the given function :
f(x)= 2x + 32/x ; when x > 0I 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 &lt; x1 or x &gt; 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].