How do you find the local max and min for F(x) = ln((x^4) + 27)F(x)=ln((x4)+27)?

2 Answers
May 3, 2016

The minimum is 3 ln 3 = 3.2963ln3=3.296, nearly, at x = 0.

Explanation:

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)

May 3, 2016

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.)