Question 203087
Here are the basics to finding the extrema (ie the max/min)


Step 1) Derive the function f(x) to get f'(x)


Since *[Tex \LARGE f(x) =\frac{x^2 + 250}{x}] this means that *[Tex \LARGE f^{\prime}(x) =\frac{d}{dx}\left(\frac{x^2 + 250}{x}\right)=\frac{x^2-250}{x^2}] or simply *[Tex \LARGE f^{\prime}(x)=\frac{x^2-250}{x^2}]



Step 2) Set the derivative function f'(x) equal to zero



*[Tex \LARGE 0=\frac{x^2-250}{x^2}]



*[Tex \LARGE x^2-250=0] Multiply both sides by {{{x^2}}} and rearrange the equation.



*[Tex \LARGE x^2=250] Add 250 to both sides.



*[Tex \LARGE x=\pm \sqrt{250}] Take the square root of both sides



*[Tex \LARGE x=\pm 5\sqrt{10}] Simplify the square root



*[Tex \LARGE x \approx \pm 15.8114] Evaluate the square root



Now because the domain is *[Tex \LARGE \left(0,\infty\right)], this means that we must reject {{{x=-15.8114}}} (as it isn't in the domain)



So the max/min occurs at the approximate value {{{x=15.8114}}} (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 {{{x=15.8114}}} 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 {{{x=15.8114}}}



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 {{{x=15.8114}}} to get 




*[Tex \LARGE f(15.8114) =\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 {{{x=5*sqrt(10)}}} to get the min value: *[Tex \LARGE f(5\sqrt{10}) =\frac{(5\sqrt{10})^2 + 250}{5\sqrt{10}}=10\sqrt{10}]. This would then mean that the absolute min of {{{f(x)=10*sqrt(10)}}} occurs at {{{x=5*sqrt(10)}}}