What are the global and local extrema of #f(x)=x^3+48/x# ?

1 Answer
Jan 22, 2018

Answer:

Local: #x = -2, 0, 2#
Global: #(-2, -32), (2, 32)#

Explanation:

To find extrema, you just find points where #f'(x) = 0# or is undefined . So:

#d/dx(x^3 + 48/x) = 0#

To make this a power rule problem, we'll rewrite #48/x# as #48x^-1#. Now:

#d/dx(x^3 + 48x^-1) = 0#

Now, we just take this derivative. We end up with:

#3x^2 - 48x^-2 = 0#

Going from negative exponents to fractions again:

#3x^2 - 48/x^2 = 0#

We can already see where one of our extrema will occur: #f'(x)# is undefined at #x = 0#, because of the #48/x^2#. Hence, that is one of our extrema.

Next, we solve for the other(s). To start, we multiply both sides by #x^2#, just to rid ourselves of the fraction:

#3x^4 - 48 = 0#
#=> x^4 - 16 = 0#
#=> x^4 = 16#
#=> x = ±2#

We have 3 places where extrema occur: #x = 0, 2, -2#. To figure out what our global (or absolute) extrema are, we plug these into the original function:

By Darshan Senthil

So, our absolute minimum is the point #(-2, -32)#, while our absolute maximum is #(2, -32)#.

Hope that helps :)