Question

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

Answer 1

The minimum is `3 ln 3 = 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)

Answer 2

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.)