If #f(x)=(x^2+36)/(2x), 1 <=x<=12#, at what point is f(x) at a minimum?

1 Answer
Jan 29, 2015

I would derive your function and do some tests on the derivative to find your points.

Remember that the derivative is a function that tells you about the INCLINATION (slope), at every point, of your original function. This inclination at point #x# can be: POSITIVE (the function goes up), NEGATIVE (the function goes down) or ZERO (neither up nor down).

Now, if you can find points where your inclination "changes" (from up to down or vice-versa, represented by ZERO inclination) you find your maxima or minima!!!
enter image source here
In your case you have:

1) I now sets equal to ZERO to find the point(s) #x# where the derivative is ZERO so that my original function is neither up or down.
I get:
#f'(x)=0# or #(2x^2-72)/(4x^2)=0#
Which gives me #x=+-6#
This means that when #x=+-6# the derivative of your function is ZERO meaning that you have a point of minima or maxima...but how do I know which?

2) I check when my derivative is bigger than zero, this means that the function goes up and I can "see" whether my points are maxima or minima!
enter image source here
You can see that the point of coordinate #x=6# is a minimum for your function (the other point #x=-6# is outside your interval of interest).

You can also plot your functon to confirm this:
graph{(x^2+36)/(2x) [-14.83, 25.17, -5.37, 14.63]}