How do you find the exact relative maximum and minimum of the function of #f(x) = 4x+6x^-1#?

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Answer 1 · 2016-05-26T12:58:37

Find the critical numbers. Test the critical numbers. Evaluate #f# at the critical numbers.

#f(x) = 4x+6x^-1# has domain #(-oo,0)uu(0,oo)#

#f'(x) = 4-6/x^2# is undefined only at #x=0# (which is not in the domain of #f#, hence is not a critical number).

#f'(x) = 0# at #x=+-sqrt(3/2)#

This is a good example to use the second derivative test for.

#f''(x) = 12/x^3# #" " # (which is continuous except at #x=0#)

So #f''(-sqrt(3/2)) < 0#, which implies that #f(-sqrt(3/2))# is a relative maximum.

And #f''(sqrt(3/2)) > 0#, which implies that #f(sqrt(3/2))# is a relative minimum.

We can save ourseles some work evaluating the function by noting that the function is odd. So #f(-sqrt(3/2)) = f(sqrt(3/2))#.

#f(sqrt(3/2)) = (4sqrt3)/sqrt2 + (6sqrt2)/sqrt3#

# = 2sqrt2sqrt3+2sqrt3sqrt2#

# = 4sqrt6#.

#f# has a relative maximum of #-4sqrt6# (at #x=-sqrt(3/2)#

and a relative minimum of #4sqrt6# (at #x=sqrt(3/2)#.