How do you find the local max and min for #F(x) = ln((x^4) + 27)#?

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Answer 1 · 2016-05-03T04:47:40

The minimum is #3 ln 3 = 3.296#, nearly, at x = 0.

Thanks to Jim for pointing out the mistake in the following step that is duly corrected now.

#F'= (4x^3)/( x^4 + 27 )# = 0, when x = 0.

#F''=(12x^2)/( x^4 + 27 )# + higher power of x = 0, at x = 0

#F'''=(24x)/( x^4 + 27 )# + higher power of x, = 0, at x = 0.

#F'''=24/( x^4 + 27 )# + higher power of x. = 24 > 0, at x = 0.

If #F, F', F'', ... F^((n))=0 and F^((n+1)) > 0#, at x = a, and n is odd, then

F(a) is a local minimum

Here a = 0, n = 3 (odd) and #F^((4))#( 0) = 24 > 0.

So, F(o) is local minimum. There is no minimum, elsewhere.. So,

#F(0) = ln 27 = ln (3^3)=3 ln 3 =3.296, nearly, is the minimum of F(x)

Answer 2 · 2016-05-03T15:00:26

Find the critical numbers:

#F'(x) = (4x^3)/(x^4+27)#

#F'(x)# is defined for all #x# in the domain of #F# and #F'(x) = 0# only at #x=0#.

The only critical number is #0#.

Test the critical numbers:

#F'(x)# is negative for #x < 0# and positive for #x > 0#,

so #F(0) = ln27# is a local minimum.

(If you prefer, you could use the second derivative test, but in this case the first derivative test is simple enough.)