SOLUTION: Find the absolute maximum and absolute minimum values of the function below. If an absolute maximum or minimum does not exist, enter NONE.
f(x) = x^2 + 250/x on the open interval
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Question 203087: Find the absolute maximum and absolute minimum values of the function below. If an absolute maximum or minimum does not exist, enter NONE.
f(x) = x^2 + 250/x on the open interval (0,infinity )
I know that the absolute max is the answer NONE
but I can not figure out the absolute min
can someone help please
thanks
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
Here are the basics to finding the extrema (ie the max/min)
Step 1) Derive the function f(x) to get f'(x)
Since  =\frac{x^2 + 250}{x})
this means that  =\frac{d}{dx}\left(\frac{x^2 + 250}{x}\right)=\frac{x^2-250}{x^2})
or simply =\frac{x^2-250}{x^2})
Step 2) Set the derivative function f'(x) equal to zero
Multiply both sides by and rearrange the equation.
Add 250 to both sides.
Take the square root of both sides
Simplify the square root
Evaluate the square root
Now because the domain is )
, this means that we must reject (as it isn't in the domain)
So the max/min occurs at the approximate value (or at the endpoints)
Now if you graph the function (or evaluate a list of x values), you'll find that there is no max value (so you are correct). So this means that the min occurs when or the min occurs at the endpoints of the interval. Because x=0 is not defined (and infinity is not a number), this means that the min must occur at
Step 3) Find the minimum value of the function f(x)
From here, simply plug in the x value that generates the min value, which in this case is to get
 =\frac{15.8114^2 + 250}{15.8114}=31.6228)
So the absolute min is approximately f(x)=31.6228 and occurs at x=15.8114
Note: if you want to keep things exact, then you can plug in to get the min value:  =\frac{(5\sqrt{10})^2 + 250}{5\sqrt{10}}=10\sqrt{10})
. This would then mean that the absolute min of occurs at
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