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

May 3, 2016

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.

$F ' = \frac{4 {x}^{3}}{{x}^{4} + 27}$ = 0, when x = 0.

$F ' ' = \frac{12 {x}^{2}}{{x}^{4} + 27}$ + higher power of x = 0, at x = 0

$F ' ' ' = \frac{24 x}{{x}^{4} + 27}$ + higher power of x, = 0, at x = 0.

$F ' ' ' = \frac{24}{{x}^{4} + 27}$ + higher power of x. = 24 > 0, at x = 0.

If $F , F ' , F ' ' , \ldots {F}^{\left(n\right)} = 0 \mathmr{and} {F}^{\left(n + 1\right)} > 0$, at x = a, and n is odd, then

F(a) is a local minimum

Here a = 0, n = 3 (odd) and ${F}^{\left(4\right)}$( 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:

$F ' \left(x\right) = \frac{4 {x}^{3}}{{x}^{4} + 27}$

$F ' \left(x\right)$ is defined for all $x$ in the domain of $F$ and $F ' \left(x\right) = 0$ only at $x = 0$.

The only critical number is $0$.

Test the critical numbers:

$F ' \left(x\right)$ is negative for $x < 0$ and positive for $x > 0$,

so $F \left(0\right) = \ln 27$ 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.)