How do you find the critical numbers of $f(x)=ln(x^4+27)$ ?

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Answer 1 · 2016-12-07T19:08:53

$x = 0$ is the only critical number.

Differentiate.

We let $y = lnu$ and $u = x^4 + 27$ . Then $dy/(du) = 1/u$ and $(du)/dx = 4x^3$ .

$dy/dx = dy/(du) xx (du)/dx$

$dy/dx = 1/u xx 4x^3$

$dy/dx = (4x^3)/(x^4 + 27)$

The function will have critical numbers when $f'(x) = 0$ or when $f'(x)$ is undefined.

Let's start by finding any asymptotes.

$x^4 + 27 =0$

$x^4 = -27$

$x = root(4)(-27)$

$:.$ The function is defined for all real values of $x$ .

Now, set the derivative to $0$ and solve.

$0 = (4x^3)/(x^4 + 27)$

$0 = 4x^3$

$x = 0$

Thus, the only critical point is at $x = 0$ .

Hopefully this helps!

How do you find the critical numbers of f(x)=ln(x^4+27)f(x)=ln(x^4+27)?