[Solved] Calculus Question
February 5th, 2020
f(x)= 2000 + 20x + 10 sqrt(x) for x>0
Show that f'(x) is a decreasing convex function of x.
My answer:
f'(x) = 20 + 5 / sqrt(x)
f”(x) = -2.5 / sqrt(x^3)
As long as x is real, f”(x) can not be greater than 0 therefore f'(x) is always a decreasing concave value.
If there is anything you think I may have missed, or ways that I can improve my answer, please let me know
Thanks in advance
thought i would be able to answer ur question since im in pre calc. lol, but i guess its quite a lot different…ahha
Hi, it is a bit tricky.
Results you have computed are not analysing the behaviour of f'(x) but simply of the original f(x).
But your task is to analyse f'(x) itself.
Hence you simply derive the f'(x), which is correctly as you have
f'(x) = 5/x^(1/2) + 20
For more clarity create a new function g(x) and let g(x) = f'(x) = 5/x^(1/2) + 20 (so that you have a ‘new function’). Now you can totally forget about the f(x) function, since you analyse g(x) which is your only subject matter now.
Allright, so now you simply search for monotony and convexity of g(x).
Verify if it is decreasing by g'(x) = -5/(2*x^(3/2)) (note that it is your f”(x) result) and just after that verify the convexity by g”(x) = 15/(4*x^(5/2)).
Since the g”(x) is certainly positive for x>0 you can conclude that the function g(x) which is a derivative of f(x) is both decreasing and convex.
Your approach was in general OK but you just needed to calculate one more grade derivative of f(x). It is a nice example indeed since it makes you think about the problem not only to calculate.
Thank you so much , that certainly cleans things up.
Thanks crackajack for reading too.
I GAVE THE SAME SUM TO THE STUDENTS LAST WEEK.
I GAVE THE SAME SUM TO THE STUDENTS LAST WEEK.
Haha maybe you’ve got a student here on W-BB
I GAVE THE SAME SUM TO THE STUDENTS LAST WEEK.
Where are you from? Is this cheating? xD
i would have taken the first derivative, and then made a quick sketch of it. You correctly took the first and second deriv, but all you really need is the first
you also have your rules that will tell you if it if concave/convex/etc, and whether it opens up/down/left/right
I GAVE THE SAME SUM TO THE STUDENTS LAST WEEK.
Where are you from? Is this cheating? xD
No dear this is not cheating, atleast you are trying to get help with something you dont understand, i think this is really a great !dea, atleast you are learning, you gave me an inspiration to start a new thread , ill ask a new question (math problem) every day and will give my rapid account for whole 24hrs who gives the correct answer, how is it ???
Keen link me the thread
Keen link me the thread
yup for sure, will start from monday..
Thank you!
Regarding Calculus, does anybody have the ebook of Salas, Hille, Etgen, Calculus: One and Several Variables, Wiley, 10th edition, 2007.
I’m looking for this desperately!
i do, will upload it tomorow and making a new thread too..
CORRECT ANSWER! Hi, it is a bit tric ky.
Results you have computed are not analysing the behaviour of f'(x) but simply of the original f(x).
But your task is to analyse f'(x) itself.
Hence you simply derive the f'(x), which is correctly as you have
f'(x) = 5/x^(1/2) + 20
For more clarity create a new function g(x) and let g(x) = f'(x) = 5/x^(1/2) + 20 (so that you have a 'new function'). Now you can totally forget about the f(x) function, since you analyse g(x) which is your only subject matter now.
Allright, so now you simply search for monotony and convexity of g(x).
Verify if it is decreasing by g'(x) = -5/(2*x^(3/2)) (note that it is your f''(x) result) and just after that verify the convexity by g''(x) = 15/(4*x^(5/2)).
Since the g''(x) is certainly positive for x>0 you can conclude that the function g(x) which is a derivative of f(x) is both decreasing and convex.
Your approach was in general OK but you just needed to calculate one more grade derivative of f(x). It is a nice example indeed since it makes you think about the problem not only to calculate.